Ch. - 8 Introduction to Trigonometry

Ch. - 8 Introduction to Trigonometry

Chapter 8 Ex.8.1 Question 1

In  \(\,\Delta  ABC,\) right-angled at \(\rm{B}\), \(\rm AB = 24 \,\rm cm,\) \(\rm BC = 7\rm\, cm\), determine:

(i) \(  \quad\text{sin}\,A,\text{cos}A \)

(ii) \(\quad \text{sin }C,\text{cos}\,C  \)

 

Solution

Video Solution

What is the known?

Two sides of a right-angled triangle \(\Delta\!\!\text{ ABC}\)

What is the unknown?

Sine and cosine of angle \( A\) and \( C\).

Reasoning:

Applying Pythagoras theorem for  \(\Delta \text{ABC,}\) we can find hypotenuse (side \(\rm AC\)). Once hypotenuse is known, we can find sine and cosine angle using trigonometric ratios.

Steps:

\(\Delta \text{ABC,}\) we obtain.

\[\begin{align}\text{A}{{\text{C}}^{\text{2}}}\, &=\,\text{A}{{\text{B}}^{\text{2}}}\text{+B}{{\text{C}}^{\text{2}}} \\ &=\,{{\text{(24}\ \text{cm)}}^{\text{2}}}\,\text{+}\,{{\text{(7}\ \text{cm)}}^{\text{2}}} \\ &=\,\text{(576+49)}\ \text{c}{{\text{m}}^{\text{2}}} \\ &=\,\text{625}\ \text{c}{{\text{m}}^{\text{2}}} \end{align}\]

\(\therefore\) Hypotenuse,

\(\text{AC}\,\text{=}\,\sqrt{\text{625}}\ \text{cm}\,\text{=}\,\text{25}\ \text{cm}\)

(i)

\[\begin{align}  \sin \text{A} & =\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{A}}{\text{hypotenuse}} \\ &=\frac{\text{BC}}{\text{AC}} \\  \sin \text{A} & =\frac{\text{7}\,\text{cm}}{\text{25}\,\text{cm}}=\frac{7}{25} \\  \sin \text{A}&=\frac{7}{25} \\  \cos \text{A} & =\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}}{\text{hypotenuse}} \\ &=\frac{\text{AB}}{\text{AC}} \\ & =\frac{24\,\text{cm}}{25\,\text{cm}}=\frac{24}{25} \\  \cos \text{A} &=\frac{24}{25} \end{align}\]

(ii)

\[\begin{align}  \text{sin}\,\text{C} & =\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{C}}{\text{hypotenuse}} \\ & =\frac{\text{AB}}{\text{AC}} \\  \text{sin}\,\text{C} &=\frac{24\,\text{cm}}{\text{25}\,\text{cm}}\text{=}\frac{24}{\text{25}} \\  \text{sin}\,\text{C} &=\frac{\text{24}}{\text{25}} \\  \text{cos}\,\text{C} &=\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{C}}{\text{hypotenuse}}\\ & =\frac{\text{BC}}{\text{AC}} \\ & =\frac{7\,\text{cm}}{\text{25}\,\text{cm}}\text{=}\frac{7}{\text{25}} \\  \text{cos}\,\text{C} &=\frac{7}{\text{25}} \end{align}\]

Chapter 8 Ex.8.1 Question 2

In the given figure, find \(\text{tan}\,\text{P }-\text{cot }\text{R}.\)

 

Solution

Video Solution

What is the known?

\(PQ = 12\,\rm{ cm}\) and \(PR = 13\,\rm {cm}.\)

What is the unknown?

One side of right-angled triangle \(\Delta PQR\)

Reasoning:

Using Pythagoras theorem, we can find the length of the third side, then the required trignometric ratios.

Steps:

Apply Pythagoras theorem for \(\Delta PQR\) we obtain:

\[\begin{align}
P{R^2}\, &= \,P{Q^2} + Q{R^2}\\
Q{R^2}\, &= \,P{R^2} - P{Q^2}\\
Q{R^2} &= {(13\rm cm)^2}\, - \,{(12\rm cm)^2}\\
Q{R^2} &= 169\,\rm c{m^2} - 144\,\rm c{m^2}\\
Q{R^2} &= 25\,\rm c{m^2}\\
QR\, &= \,5\,\rm cm
\end{align}\]

\[\begin{align}  \text{tan}\,\text{P} &= \frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{P}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{P}}\\ &= \frac{\text{QR}}{\text{PQ}} \\ & = \frac{\text{5}\,\text{cm}}{\text{12}\,\text{cm}} \\ \text{tan}\,\text{P} & = \frac{\text{5}}{\text{12}} \end{align}\]

\[\begin{align} \rm{cot}\,{R} &= \frac{ \text{side adjacent to}\ \angle {R}}{\text{side opposite to}\ \angle {R}} \\ & = \frac{{QR}}{{PQ}} \\ & =\frac{{5}\,\rm{cm}}{{12}\,\rm{cm}} \\ \rm {cot}\,{R} &= \frac{{5}}{{12}} \end{align}\]

\[\begin{align} \rm {\tan} \,{P}- {{cot}}\,{R} & = \frac{{5}}{{12}}-\frac{{5}}{{12}} \\ \rm{tan}\,P- {cot}\,{R} &= {0}\end{align}\]

Chapter 8 Ex.8.1 Question 3

If \(\begin{align}\text{sin A}=\frac{3}{4}\end{align}\) calculate \(\rm{cos\,A}\) and \(\rm {tan\,A}.\)

 

Solution

Video Solution

What is the known?

Sine of \(\angle \text{A}\) .

What is the unknown?

Cosine and tangent of \(\angle \text{A}\)

Reasoning:

Using sin A, we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

Let \(\Delta \text{ABC}\) be a right-angled triangle, right angled at point \(\rm{B} \).

Given that:

\[\begin{align} \Rightarrow {\sin A}&={\frac{3}{4}} \\ {\frac{B C}{A C}}&={\frac{3}{4}}\end{align}\]

Let \(\rm{BC}\) be \(3k\). Therefore, \(\rm{AC}\) will be \(4k\) where \(k \) is a positive integer.

Applying Pythagoras theorem for \(\Delta \,\rm ABC,\)  we obtain:

\[\begin{align}
A{C^2} &= A{B^2}\, + \,B{C^2}\\
A{B^2} &= A{C^2} - \,B{C^2}\\
A{B^{2\,}} &= {(4\,k)^2} - \,{(3\,k)^2}\\
A{B^2} &= 16{k^2} - 9\,{k^2}\\
A{B^2}\, &= \,7\,{k^2}\\
AB\, &= \sqrt {7\,} k
\end{align}\]

\[\begin{align} \,\text{cosA}\,&=\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}}{\text{hypotenuse}} \\ & =\frac{\text{AB}}{\text{AC}}\\ & =\frac{\sqrt{\text{7}}k}{\text{4}\,k} \\&=\frac{\sqrt{\text{7}}}{\text{4}} \end{align}\]

\[\begin{align} \text{tan}\,\text{A}&=\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{A}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}} \\ & =\frac{\text{BC}}{\text{AB}}\,\,\,\text{=}\frac{\text{3}\,k}{\sqrt{\text{7}}\,k} \\ &=\frac{\text{3}}{\sqrt{\text{7}}} \end{align}\]

Thus,

\[\begin{align}\,\text{cos}\,\text{A=}\frac{\sqrt{\text{7}}}{\text{4}}\ \ \text{and}\ \text{tan}\ \text{A=}\frac{\text{3}}{\sqrt{\text{7}}}\end{align}\]

Chapter 8 Ex.8.1 Question 4

Given \(15 \,\rm{cot A} = 8,\) find \(\rm{sin\, A}\) and \(\rm{sec\,A.}\)

 

Solution

Video Solution

What is the known?

Cotangent of \(\angle A \)

What is the unknown?

Sine and Secant of \(\angle A \) .

Reasoning:

Using cot A, we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

Let us consider a right-angled \(\Delta \rm{ABC},\) right angled at \(\rm{B.}\)

\[\begin{align}\text{cot}\,\text{A}\, \text{=}\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}}{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{A}}\text{=}\frac{\text{AB}}{\text{BC}}\end{align}\]

It is given that

\[\begin{align}\text{cotA}\,\text{=}\,\frac{\text{8}}{\text{15}}\Rightarrow \frac{\text{AB}}{\text{BC}}\,\text{=}\,\frac{\text{8}}{\text{15}}\end{align}\]

Let \(\rm{AB}\) be \(8\,k.\) Therefore, \(\rm{BC}\) will be \(15\,k\) where \(k\) is a positive integer.

Apply Pythagoras theorem in \(\text{ }\!\!\Delta\!\!\text{ }\,\text{ABC,}\) we obtain.

\[\begin{align}
A{C^2}\, &= \,A{B^2} + B{C^2}\\
A{C^2}\, &= \,{(8k)^2} + {(15k)^2}\\
A{C^2}\, &= \,64{k^2} + 225{k^2}\\
A{C^2}\, &= \,289{k^2}\\
\,\;\,AC\,\, &= \,17k
\end{align}\]

\[\begin{align}
{\rm{sin}}\,{\rm{A}} &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle {\rm{A}}}}{{{\rm{hypotenuse}}}} \\ & = \frac{{BC}}{{AC}} = \frac{{15\,k}}{{17\,k}}\\
& = \frac{{{\rm{15}}}}{{{\rm{17}}}}\\
{\rm{sec}}\,{\rm{A}} & = \frac{{{\rm{hypotenuse}}}}{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle {\rm{A}}}}\\ & = \frac{{AC}}{{AB}} = \frac{{17\,k}}{{8\,k}}\\
 &= \frac{{{\rm{17}}}}{{\rm{8}}}
\end{align}\]

Thus,

\[\begin{align}\text{sin}\,\text{A}\,\text{=}\,\frac{\text{15}}{\text{17}}\ \ \text{and}\ \ \text{sec}\,\text{A}\,\text{=}\,\frac{\text{17}}{\text{8}}\end{align}\]

Chapter 8 Ex.8.1 Question 5

Given \(\begin{align}\text{sec}\,\theta =\frac{13}{12},\end{align}\) calculate all other trigonometric ratios.

 

Solution

Video Solution

What is the known?

\(\text{Secant}\ \rm{of} \,\theta \)

What is the unknown?

Other trigonometric ratios.

Reasoning:

Using \(\rm Sec\;\theta\) , we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

Let \(\Delta \rm{ABC}\) be a right-angled triangle, right angled at point \(\rm{B.}\)

It is given that:

\[\begin{align}\text{sec}\ \text{ }\!\!\theta\!\!\text{ }\,&=\frac{\text{hypotenuse}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle \theta }\\ &=\frac{\text{AC}}{\text{AB}} \\ & =\frac{\text{13}}{\text{12}}\end{align}\]

Let \(AC = 13\,k\) and \(BC = 12\,k\) where \(k \) is a positive integer.

Apply Pythagoras theorem in \(\begin{align}\triangle \rm{ABC}\end{align}\) we obtain:

\[\begin{align}  A{C^2} &= A{B^2} + B{C^2}\\
B{C^2} &= A{C^2} - A{B^2}\\
B{C^2} &= {(13\,k)^2} - {(12\,k)^2}\\
B{C^2} &= 169\,{k^2} - 144\,{k^2}\\
B{C^2} &= 25\,{k^2}\\
BC &= 5\,k\end{align}\] 

\[\begin{align} \\{\sin \theta}&=\frac{\text { side opposite to } \angle \theta}{\text { hypotenuse }} \\ & =\frac{\mathrm{BC}}{\mathrm{AC}} \\ & =\frac{5}{13} \\ {\cos \theta}&=\frac{\text { side adjacent to } \angle \theta}{\text { hypotenuse }} \\ &=\frac{\mathrm{AB}}{\mathrm{AC}} \\ & =\frac{12}{13} \\ {\tan \theta}&=\frac{\text { side opposite to } \angle \theta}{\text { side adjacent to } \angle \theta} \\ &=\frac{\mathrm{BC}}{\mathrm{AB}} \\ & =\frac{5}{12} \\ {\cot \theta}&=\frac{\text { side adjacent to } \angle \theta}{\text { side opposite to } \angle \theta} \\ &=\frac{\mathrm{AB}}{\mathrm{BC}} \\ & =\frac{12}{5}\\ \text{cosec}\,\text{ }\!\!\theta\!\! &=\frac{\text{hypotenuse}}{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{ }\!\!\theta\!\!\text{ }} \\ & =\frac{\text{AC}}{\text{BC}} \\ & = \frac{\text{13}}{\text{5}}\end{align}\]

Chapter 8 Ex.8.1 Question 6

If \(\,\,\angle {A}\) and \(\,\,\angle {B}\) are acute angles such that \(\,\,\text{cos}\,{A}=\text{cos}\,{B},\) then show that \(\angle {A}=\angle \,{B}\,{.}\)

 

Solution

Video Solution

What is the known?

\(\angle {A}\) and \(\,\,\angle {B}\) are acute angles and \(\begin{align}\text{cos}\, A  = \text{cos}\,  B\end{align}\)

What is the unknown?

To show that \(\angle A= \angle B\)

Reasoning:

Using \(\text{cos} \,A\) and \(\text{cos}\, B,\) we can find the ratio of the length of two sides of the right-angled triangle with respective angles. Then compare both the ratios.

Steps:

In the right-angled triangle \(ABC, \,\)\(\,\angle A\)  and  \(\angle B \) are acute angles and \(\angle C\) is right angle.

\[\begin{align} & \text{cos}\,{A}\,{=}\,\frac{\text{side adjacent to}\ \angle {A}}{\text{hypotenuse}}{=}\frac{{AC}}{{AB}} \\ & \text{cos}\,{B}\,{=}\,\frac{\text{side adjacent to}\,\angle {B}}{\text{hypotenuse}}{=}\,\frac{{BC}}{{AB}} \end{align}\]

Given that \(\begin{align} \;\text{cos} \,A&=\,\text{cos} B \end{align}\)

\(\begin{align} \text{Therefore, }\frac{{AC}}{{AB}}\,&=\,\frac{{BC}}{{AB}} \\ {AC}\,&=\,{BC}\end{align}\)

Hence, \(\angle A=\angle B\,,\) (angles opposite to equal sides of triangle are equal)

Alternatively,

Reasoning:

Using \(\text{ cos} \,A\) and \(\text{cos}\, B,\) we can find the ratio of the length of two sides of the right-angled triangle with respective angles. Then by using Pythagoras theorem, relation between the sides

Let us consider a triangle \(ABC\) in which \(\begin{align}{CO} \perp {AB}\end{align}\)

It is given that

\[\begin{align} \text{cos}\,{A}\,&=\, \text{cos}\,{B} \\ \frac{{AO}}{{AC}}&=\frac{{BO}}{{BC}} \\ \frac{{AO}}{{BO}}\,&=\,\frac{{AC}}{{BC}}\\ \\ \text{ Let }\; \frac{ {A O}}{{B O}}&=\frac{{A C}}{{B C}}={k} \\ { A O}&={k B O}\;\dots\rm{(i)} \\ {A C}&={k B C}\;\dots \rm{(ii)} \end{align}\]

By applying Pythagoras theorem in  \(\Delta {CAO}\) and \(\Delta {CBO},\) we get.

\[\begin{align}  {A}{{{C}}^{{2}}} & ={A}{{{O}}^{{2}}}{+C}{{{O}}^{{2}}}[\text{from} \;\Delta\;{ CAO}] \\  {C}{{{O}}^{{2}}}&={A}{{{C}}^{{2}}}-{A}{{{O}}^{{2}}}\cdots \left(\rm {iii} \right) \\  {B}{{{C}}^{{2}}}&={B}{{{D}}^{{2}}}{+C}{{{O}}^{{2}}}[\text {from}\;\Delta\,{ CBO}] \\  {C}{{{O}}^{{2}}}&={B}{{{C}}^{{2}}}\,-{B}{{{O}}^{{2}}}\cdots \left( \rm{iv} \right) \\ \end{align}\]

From equation (iii) and equation (iv), we get

\[\begin{align} {A}{{{C}}^{{2}}}-{A}{{{O}}^{{2}}}&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\ \ {{{(kBC)}}^{{2}}}-{{{(kBO)}}^{{2}}}&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\  {{k}^{{2}}}{B}{{{C}}^{{2}}}-{{k}^{{2}}}{B}{{{O}}^{{2}}}\,&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\  {{k}^{{2}}}{(B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}}{)}\,&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\  {{k}^{{2}}}\,&=\,\frac{{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}}}{{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}}} \\ &=1 \\k\,&=1 \end{align}\]

Putting this value in equation (ii) we obtain

\[{AC}\,{=}\,{BC}\]

\({\angle A=\angle B}\) (angles opposite to equal sides of triangle are equal.)

Chapter 8 Ex.8.1 Question 7

If \(\cot \theta  = \frac{7}{8}\), evaluate:

(i) \(\begin{align}\,\,{\rm{ }}\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}}\end{align}\), 

(ii) \(\begin{align} {\rm{ co}}{{\rm{t}}^2}\theta \end{align} \)

 

Solution

Video Solution

 

What is the known?

\(\cot \theta  = \frac{7}{8}\)

What is the unknown?

Value of \(\begin{align} \left( \rm{i} \right)\,\,{\rm{  }}\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}} \end{align}\), and   \(\begin{align} \left( \rm{ii} \right){\rm{  co}}{{\rm{t}}^2}\theta \end{align} \)

Reasoning:

Using \(\cot \theta  = \frac{7}{8}\), we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

Let \({\rm{\Delta }}\,{{ABC}}\), in which angle \(B\) is right angle.

\[\begin{align} \cot \theta  &= \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\theta }}{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\theta }}\\ &= \frac{{AB}}{{BC}}\\&= \frac{7}{8} \end{align}\]

Let \(AB = 7k\)and \(BC = 8k\), where \(k\) is a positive integer.

By applying Pythagoras theorem in we get.

\[\begin{align}{\rm A}{C^2}\, & = {\rm A}{{\rm B}^2} + {\rm B}{C^2}\\ &= {(7k)^2} + {(8k)^2}\\ &= 49{k^2} + 64{k^2}\\ &= 113{k^2}\\
AC\, &= \,\sqrt {113k{\,^2}} \\ &= \,\sqrt {113} k\end{align}\]

Therefore,

\[\begin{align}{\rm{sin }}\theta &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\theta }}{{{\rm{hypotenuse}}}} = \frac{{BC}}{{AC}}\\ &= \frac{{8k}}{{\sqrt {113} k}} = \frac{8}{{\sqrt {113} }}\\\cos \theta &= \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\theta }}{{{\rm{hypotenuse}}}} = \frac{{AB}}{{AC}} \\ & = \frac{{7k}}{{\sqrt {113} k}} = \frac{7}{{\sqrt {113} }}\end{align}\]

\(\begin{align}\left(\rm{i} \right)\,\,{\rm{  }}\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}} \end{align}\)

\[\begin{align}&\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}} \\ &= \frac{{1 - {{\sin }^2}\theta }}{{1 - {{\cos }^2}\theta }} \\ & \left[ \because {\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)} \right]\\ &= \frac{{1 - {{\left( {\frac{8}{{\sqrt {113} }}} \right)}^2}}}{{1 - {{\left( {\frac{7}{{\sqrt {113} }}} \right)}^2}}}\\ &= \frac{{1 - \frac{{64}}{{113}}}}{{1 - \frac{{49}}{{113}}}}\\
 &= \frac{{\frac{{49}}{{113}}}}{{\frac{{64}}{{113}}}}\\ &= \frac{{49}}{{64}}\end{align}\]

\(\left( \rm{ii} \right){\rm{  co}}{{\rm{t}}^2}\theta \)

\[\begin{align}{\rm{co}}{{\rm{t}}^2}\theta  &= {\left( {\frac{7}{8}} \right)^2}\\ &= \frac{{49}}{{64}}\end{align}\]

Chapter 8 Ex.8.1 Question 8

If \(3 \;cot A = 4\), check whether \(\begin{align} \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A \end{align}\) or not.

 

Solution

Video Solution

 

What is known?

Cotangent of angle \(A\)

What is unknown?

Whether \(\begin{align} \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A \end{align}\)

Reasoning:

Using \(3\cot {{A}} = 4\), we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

\[\begin{align}{\rm{3}}\,{\rm{cot}}\,{{A}}\,\, &= \,\,{{4}}\\{\rm{cot}}\,{{A}}\,\, &= \,\,\frac{{\rm{4}}}{{\rm{3}}}\end{align}\]

Let , in which angle \(B\) is right angle.

\[ \begin{align} \cot A & = \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle A}} \\ & = \frac{{AB}}{{BC}} = \frac{4}{3} \end{align} \]

Let \(AB = 4k\;{\rm{ }}{\rm{and }}\;BC = 3k\) where \(k\) is a positive integer.

By applying Pythagoras theorem in \(\Delta ABC\) we get.

\[\begin{align}{A}{C^2}\, & = {A}{{B}^2} + {B}{C^2}\\ &= {(4k)^2} + {(3k)^2}\\ &= 16{k^2} + 9{k^2}\\ &= 25{k^2}\\{{AC}}\,& = \,\sqrt {{{25k}}{\,^{{2}}}} \\&={{5k}}\end{align}\]

Therefore,

\[ \begin{align} {\rm{tan}}\,A &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle A}}{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}\\ & = \frac{{BC}}{{AB}} \\ & = \frac{{3k}}{{4k}}  \\ &= \frac{3}{4} \end{align} \]

\[\begin{align}{\text{sin}} A &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle {{A}}}}{{{\rm{hypotenuse}}}}{\rm{ = }}\frac{{{{BC}}}}{{{{AC}}}}\\&=\frac{{{\rm{3k}}}}{{{\rm{5k}}}}= \frac{{\rm{3}}}{{\rm{5}}} \end{align}\]

\[ \begin{align} \cos A & = \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}{{{\rm{hypotenuse}}}} \\ & = \frac{{AB}}{{AC}} \\ & = \frac{{4k}}{{5k}} \\ & = \frac{4}{5} \end{align} \]

\[\begin{align}L.H.S &= \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}\\ &= \frac{{1 - {{\left( {\frac{3}{4}} \right)}^2}}}{{1 + {{\left( {\frac{3}{4}} \right)}^2}}}\\ &= \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}}\\& = \frac{{16 - 9}}{{16 + 9}}\\ &= \frac{7}{{25}}\end{align}\]

\[\begin{align}R.H.S\,\, &= {\cos ^2}A - {\sin ^2}A\\&= {\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2}\\ &= \frac{{16}}{{25}} - \frac{9}{{25}}\\&= \frac{{16 - 9}}{{25}}\\& = \frac{7}{{25}}\end{align}\]

Therefore, \(\begin{align}\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A \end{align}\)

Chapter 8 Ex.8.1 Question 9

In the \(\Delta {ABC}\) right-angled at \({B}\), if \(\begin{align}\text{tan}\,A=\frac{\text{1}}{\sqrt{\text{3}}}\end{align}\) find the value of:

(i) \( \quad \text{sin }A \text{ cos }{C }+\text{ }\text{cos }A \;\text{sin }{C } \) 

(ii) \( \quad \text{cos }{A}\;\text{cos }{C }-\text{ }\text{sin }{A }\text{ sin }{C } \)

 

Solution

Video Solution

Reasoning:

Using \(\tan {{A}} = \frac{1}{{\sqrt 3 }}\), we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

(i) Let \(\begin{align}\triangle {ABC}\end{align}\) be a right-angled triangle \(\begin{align}\text{tan}\,A=\frac{\text{1}}{\sqrt{\text{3}}}\end{align}\)

\[\begin{align}\text{tan}\,A\,& =\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle A}{\text{side}\ \text{adjacent}\ \text{to}\ \angle A} \\ & =\frac{BC}{AB}=\frac{1}{\sqrt{3}}\end{align}\]

Let  \({BC} = {k}\) and \({AB} = \sqrt{3\,}{k}\)  where \(k\) is a positive real number.

By applying Pythagoras theorem for \(\text{ }\!\!\Delta\!\!\text{ }\,ABC\)

\[\begin{align} {AC}^{2} &={AB}^{2}+{BC}^{2} \\ &=(\sqrt{3} {k})+({k})^{2} \\ &=3 {k}^{2}+{k}^{2} \\ &=4 {k}^{2} \\ {AC} &=\sqrt{4 {k}^{2}} \\ &=2 {k} \end{align}\]

Therefore,

\[\begin{align}\sin A&=\frac{\text { side opposite to } \angle A}{\text { hypotenuse }} \\ &=\frac{B C}{A C}=\frac{1}{2} \\ \cos A&=\frac{\text { side adjacent to } \angle A}{\text { hypotenuse }} \\&=\frac{A B}{A C}=\frac{\sqrt{3}}{2} \\ \sin C&=\frac{\text { side opposite to } \angle C}{\text { hypotenuse }} \\ &=\frac{A B}{A C}=\frac{\sqrt{3}}{2} \\ \cos C&=\frac{\text { side adjacent to } \angle C}{\text { hypotenuse }} \\ & =\frac{B C}{A C}=\frac{1}{2}\end{align}\]

\(\text{(i)  sin }A\text{ cos }C+\text{cos }A\text{ sin }C\)

(By substituting the values of the trigonometric functions in the above equation.)

\[\begin{align} & \text{ sin }A\text{ cos }C+\text{cos }A\text{ sin }C \\ &=  \left( \frac{1}{2} \right)\! \left( \frac{1}{2} \right)\!+\!\left( \frac{\sqrt{3}}{2} \right)\! \left( \frac{\sqrt{3}}{2} \right)   \\ & =\frac{1}{4}+\frac{3}{4} \\ & =\frac{1+3}{4} \\ & =\frac{4}{4} \\ & =1 \end{align}\]

\(\text{(ii)}\cos \,A\,\cos \,C-\sin \,A\,\sin \,C\)

By substituting the values of the trigonometric functions in the above equation.

(ii)

\[\begin{align} & \cos \,A\,\cos \,C-\sin \,A\,\sin \,C \\ &=   \left( \frac{\sqrt{3}}{2} \right)\! \left( \frac{1}{2} \right)  \!-\!\left( \frac{1}{2} \right)\! \left( \frac{\sqrt{3}}{2} \right)  \\ & =\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4} \\ & =0 \end{align}\]

Chapter 8 Ex.8.1 Question 10

In \(\,\,\Delta PQR,\) right-angled at \(\rm{Q,}\) \(PR+QR=25 \text{cm}\) and \({PQ}=5 \text{cm}.\) Determine the values of \(\text{sin} \,P, \text{cos} \,P\) and \(\text{tan}\, P.\)

 

Solution

Video Solution

Reasoning:

Using Pythagoras theorem, we can find the length of the all three sides. Then the required trignometric ratios

Steps:

Given, \(\text{ }\!\!\Delta ABC\) be a right angle at \({Q.} \)

\[\begin{align} & {PQ=5}\,\text{cm} \\ & {PR+QR=25} \\ \end{align}\]

Let \({PR}= x \, \text{cm}\)

Therefore,

\[\begin{align} QR\, &=25\,\text{cm}- PR \\ &=25\,\text{cm}-x\,\text{cm} \end{align}\]

By applying Pythagoras theorem for \(\begin{align}\Delta {PQR}\end{align}\) we obtain.

\[\begin{align} P{{R}^{2}}&=P{{Q}^{2}}+Q{{R}^{2}} \\ {{x}^{2}}&={{(5)}^{2}}+{{(25-x)}^{2}} \\ {{x}^{2}}&=25+625-50x+{{x}^{2}} \\ 50x&=650 \\ x&=\frac{650}{50} \\ &=130\ \rm{ m} \end{align}\]

Therefore,

\[\begin{align} {PR} &=13 \rm{cm} \\ {QR} &=(25-13) \rm{cm} \\ &=12 \rm{cm} \end{align}\]

By substituting the values obtained above in the trigonometric functions below.

\[\begin{align}\sin {P}&=\frac{\text { side opposite to } \angle {P}}{\text { hypotenuse }}\\ &=\frac{{QR}}{{PR}}=\frac{12}{13} \\\\ \cos {P} & =\frac{\text { side adjacent to } \angle {P}}{\text { hypotenuse }} \\ & =\frac{{PQ}}{{PR}}=\frac{5}{13} \\\\ \tan {P} & =\frac{\text { side opposite to } \angle {P}}{\text { side adjacent to } \angle {P}} \\ & =\frac{{QR}}{{PQ}}=\frac{12}{5}\end{align}\]

Chapter 8 Ex.8.1 Question 11

State whether the following are true or false. Justify your answer.

(i) The value of \(\text{tan}\,A\) is always less than \(1. \)

(ii) \(\begin{align}\text{sec}A=\frac{\text{12}}{\text{5}}\end{align}\) for some value of \(\angle A\).

(iii) \(\text{cos}\,A\) is the abbreviation used for the cosecant of \({\angle A}\)

(iv) \(\text{cot}\, A\) is the product of \(\rm{cot} \) and \({A.}\)

(v) \(\begin{align}\sin \,\theta =\frac{4}{3},\end{align}\) for some \(\begin{align}\angle \theta\end{align}\)

 

Solution

Video Solution

Steps:

(i) False, because sides of a right-angled triangle may have any length. So \(\text{tan}\,A\) may have any value.

(ii) \(\begin{align}\text{sec}\,{A=}\,\frac{\text{hypotenuse}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle {A}}\end{align}\)

As hypotenuse is largest side, the ratio on RHS will be greater than 1. Hence \(\begin{align}\text{sec }\text{A}>1.\end{align}\) Thus, the given statement is true.

(iii) Abbreviation used for cosecant of \( \angle A\) is \(\text{cosec}\,A\) and \(\text{cos}\,A\) is the abbreviation used for cosine of \(\rm \angle A\). Hence the given statement is false.

(iv) \(\text{cot}\,A\) is not the product of \(\rm{cot}\) and \({A.}\) It is the cotangent of \(\rm \angle A\) . Hence, the given statement is false.

(v)\(\begin{align}\text{Sin}\,\theta =\frac{4}{3}\end{align}\)

We know that in a right-angled triangle,

\[\begin{align}\text{Sin}\,\theta =\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \theta }{\text{hypotenuse}}\end{align}\]

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Also, the value of Sine should be less than \(1.\) Therefore, such value of \(\rm{Sin\, \theta}\) is not possible. Hence the given statement is false.

At the beginning of the chapter 8, it is mentioned that trigonometry is the study of relationships between the sides and angles of a triangle. Other concepts covered are the trigonometric ratios of the angle and trigonometric identities. Specifically, some of the trigonometric identities explained in this chapter are - sine, cosine, tangent, cosecant, secant and cotangent. For the second exercise, trigonometric ratios of some specific angles are presented. Next, for the third exercise, the important concept of trigonometric ratios for complementary angles is taken up. Finally, the chapter covers different trigonometric identities and their applications.

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