Ex.8.4 Q4 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

Choose the correct option. Justify your choice.

(i)  \( \text{ 9 sec 2A - 9 tan 2A}  = \_\_\_\_\_\_\_\_\_\_\_\_\_ \)

(A) \(1 \) 

(B) \( 9\)

(C) \(8\) 

(D) \(0\)

(ii) \(\left( \text{1 + tan } \theta\text{ + sec }\theta \right) \left( \text{1 + cot } \theta -\text{cosec }\theta \right) = \_\_\_\_\_\_\_\_\_\_\_\_\_ \)

(A) \(0\)

(B) \(1\) 

(C) \(2\) 

(D) \(-1\)

(iii) \(\left( \text{sec A + tan A} \right) \left( \text{1 -sin A} \right) =\_\_\_\_\_\_\_\_\_\) 

(A) \(\text{ sec A }\)

(B) \(\text{ sin A } \)

(C) \( \text{ cosec A }\)

(D) \( \text{ cos A}\)

(iv) \(\frac{\text{1+ta}{{\text{n}}^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}} = \_\_\_\_\_ \) 

(A) \( \text{ sec2A}\) 

(B) \(1\) 

(C) \(\text{cot2A} \)

(D) \(\text{tan2A}\)

Text Solution

 

Reasoning:

\[\begin{align}{\sin ^{2} A+\cos ^{2} A=1} \\ {\text{cosec} ^{2} A=1+\cot ^{2} A} \\ {\sec ^{2} A=1+\tan ^{2} A}\end{align}\]

Steps:

(i)

\[\begin{align} & \text{ 9 se}{{\text{c}}^{\text{2}}}\text{A }-\text{ 9 ta}{{\text{n}}^{\text{2}}}\text{A} \\ \\ &=9\text{ (se}{{\text{c}}^{\text{2}}}\text{A}-\text{ta}{{\text{n}}^{\text{2}}}\text{A) } \\& =\text{9 }\left( \text{1} \right) \\ & \begin{bmatrix} \text{By the identity,} \\ 1+ \sec ^2 A = \tan ^2 A \\ \Rightarrow \sec ^2 A - \tan^2A=1 \end{bmatrix}  \\ & = 9\end{align}\]

(ii)

\[\begin{align} \left( \text{1 + tan }\theta\text{ + sec } \theta \right)  \left( \text{1 + cot }\theta\text{ - cosec }\theta \right) \ldots (1) \end{align}\]

We know that the trigonometric functions,

\[\begin{align} & \tan (x)=\frac{\sin (x)}{\cos (x)} \\ & \cot (x)=\frac{\cos (x)}{\sin (x)}=\frac{1}{\tan (x)} \end{align}\]

And

\[\begin{align} \sec (x)&=\frac{1}{\cos (x)} \\ {cosec}(x)&=\frac{1}{\sin (x)} \end{align}\]

By substituting the above function in Equation \((1)\),

\[\begin{align}& = \begin{bmatrix} \left( 1+ \frac{\text{sin }\theta }{\text{cos }\theta } + \frac{1}{\text{cos }\theta } \right)  \left( 1+ \frac{\text{cos }\theta }{\text{sin }\theta } - \frac{1}{\text{sin }\theta} \right) \end{bmatrix} \\ & = \begin{bmatrix} \left( \frac{\text{cos }\theta\text{ +sin }\theta+1}{\text{cos }\theta} \right)  \left( \frac{\text{sin }\theta\text{ +cos }\theta -1}{\text{sin }\theta} \right)\end{bmatrix}\\& \text{(By taking LCM and multiplying)}\\\\& = \frac{{{\text{(sin }\theta +\text{cos} \theta )}^2}-(1)^2}{\text{sin }\theta \text{cos }\theta}\\& \text{(Using )}\\ \\&= \frac{ \begin{Bmatrix}\text{sin}^2 \theta \text{ +co}{{\text{s}}^2} \theta \\ \text{ +2sin }\theta\text{ cos }\theta -1 \end{Bmatrix} }{\text{sin }\theta\text{ cos }\theta} \\ & = \frac{\text{1+2sin }\theta\text{ cos }\theta -1 }{\text{sin }\theta\text{ cos}\theta}\\&\text{(Using identify)}\\\\&=\frac{\text{2sin }\theta\text{ cos }\theta\text{ }}{\text{sin }\theta\text{ cos }\theta\text{ }}\text{= 2}\end{align}\]

Hence, option (C) is correct.

(iii)

\(\begin{align}\left( \text{sec A + tan A} \right)  \left( 1 - \text{sin A} \right) \ldots \left( 1 \right)\end{align}\)

We know that the trigonometric functions,

\[\begin{align}\tan (x)=\frac{\sin (x)}{\cos (x)}\end{align}\]

And

\[\begin{align}\sec (x)=\frac{1}{\cos (x)}\end{align}\]

By substituting the above function in Equation \((1)\),

\[\begin{align}  &= \begin{bmatrix} \left( \frac{1}{\text{cosA}} + \frac{\text{sinA}}{\text{cosA}} \right)  (1-\text{sinA)} \end{bmatrix} \\ & =\left( \frac{\text{1+sinA}}{\text{cosA}} \right) (1-\text{sinA)} \\ & = \frac{1-\text{si}{{\text{n}}^{2}}\text{A}}{\text{cosA}} \\ & = \frac{\text{co}{{\text{s}}^{2}}\text{A}}{\text{cosA}}   \\ & \begin{bmatrix} \text{By identity}\\\sin^2 \theta+ \cos^{2} \theta=1  \\ \Rightarrow  1- {\rm{sin^2}} θ = {\rm{cos^2}} θ \end{bmatrix} \\ &= \cos A\end{align}\]
Hence, option (D) is correct. 

(iv) \(\begin{align}\frac{1+{{\tan }^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}}\end{align}\)

We know that the trigonometric functions,

\[\begin{align} \tan (x)&=\frac{\sin (x)}{\cos (x)} \\ \cot (x)&=\frac{\cos (x)}{\sin (x)} \\ & =\frac{1}{\tan (x)} \end{align}\]

By substituting the above function in Equation (\(1\)),

\[\begin{align}\frac{{{\rm{1 + ta}}{{\rm{n}}^{{2}}}{{A}}}}{{{\rm{1 + co}}{{\rm{t}}^{{2}}}{{A}}}}&= \frac{{{{1 + }}\frac{{{\rm{sin}}^2{{A}}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}}}{{{{1 + }}\frac{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}{{{\rm{si}}{{\rm{n}}^{{2}}}{\rm{A}}}}}}\\ &= \frac{{\frac{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}} + {\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}}}{{\frac{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}} + {\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}}}\\&=\frac{{\frac{{{1}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}}}{{\frac{1}{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}}}\\&=\frac{{{\rm{si}}{{\rm{n}}^{{2}}}{{A}}}}{{{\rm{co}}{{\rm{s}}^{{2}}}{{A}}}}\\&= \rm{tan}^2 A\end{align}\]

Hence, option (\(D\)) is correct

  
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