Ex.8.1 Q9 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

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) \(\begin{align}\text{sin }A \text{ cos }{C }+\text{ }\text{cos }A \;\text{sin }{C } \end{align}\) 
(ii) \(\begin{align}\text{cos }{A}\;\text{cos }{C }-\text{ }\text{sin }{A }\text{ sin }{C } \\ \end{align}\)

Text 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.

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