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

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Question

Choose the correct option. Justify your choice.

\(\begin{align} & \left( \text{i} \right)\text{ 9 sec 2A - 9 tan 2A = }\_\_\_\_\_\_\_\_\_ \\ & \left( \text{A} \right)\text{ 1 } \\ & \left( \text{B} \right)\text{ 9} \\ & \left( \text{C} \right)\text{ 8} \\ & \left( \text{D} \right)\text{ 0} \\ \end{align}\)

\(\begin{align} & \left( \text{ii} \right)\text{ }\left( \text{1 + tan }\!\!\;\theta\!\!\text{ + sec }\;\!\!\theta\!\!\text{ } \right)\text{ }\left( \text{1 + cot }\;\!\!\theta\!\!\text{ }-\text{cosec }\;\!\!\theta\!\!\text{ } \right) \\ & \left( \text{A} \right)\text{ 0 } \\ & \left( \text{B} \right)\text{ 1 } \\ & \left( \text{C} \right)\text{ 2 } \\ & \left( \text{D} \right)\text{ }-\text{1} \\ \end{align}\)

\(\begin{align} & \left( \text{iii} \right)\text{ }\left( \text{sec A + tan A} \right)\text{ }\left( \text{1 -- sin A} \right)\text{ = }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } \\ & \left( \text{A} \right)\text{ sec A } \\ & \left( \text{B} \right)\text{ sin A } \\ & \left( \text{C} \right)\text{ cosec A } \\ & \left( \text{D} \right)\text{ cos A} \\ \end{align}\)

\(\begin{align} \left( \text{iv} \right)\frac{\text{1+ta}{{\text{n}}^{2}}\text{A}}{1+{{\cot }^{2}}\text{A}} \\ \left( \text{A} \right)\,\text{ sec2A} \\ \left( \text{B} \right)\,\,-\text{1} \\ \left( \text{C} \right)\,\,\text{cot2A} \\ \left( \text{D} \right)\,\text{tan2A} \\ \end{align}\)

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

(ii)     \(\begin{align}\text{ }\left( \text{1 + tan }\theta\text{ + sec }\!\;\!\theta\!\!\text{ } \right)\text{ }\left( \text{1 + cot }\theta\text{ - cosec }\theta\text{ } \right)\text{ }\ldots \ldots \ldots\end{align}\) Equation (1)

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

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

By substituting the above function in Equation (1),

\(\begin{align}  \text{=}\left( \text{1+}\frac{\text{sin }\theta\text{ }}{\text{cos }\theta\text{ }}\text{+}\frac{\text{1}}{\text{cos }\theta\text{ }} \right)\left( \text{1+}\frac{\text{cos }\theta\text{ }}{\text{sin }\theta\text{ }}\text{-}\frac{\text{1}}{\text{sin }\theta\text{ }} \right)\end{align}\)
\(\begin{align}  & \text{=}\left( \frac{\text{cos }\theta\text{ +sin }\theta\text{ +1}}{\text{cos }\theta\text{ }} \right)\left( \frac{\text{sin }\theta\text{ +cos }\theta\text{ -1}}{\text{sin }\theta\text{ }} \right)  \end{align}\) (By taking LCM and multiplying) 
\(\begin{align}  & \text{=}\frac{{{\text{(sin }\theta\text{ +cos }\theta\text{ )}}^{\text{2}}}\text{-(1}{{\text{)}}^{\text{2}}}}{\text{sin }\theta\text{ }\,\,\text{cos }\theta\text{ }}\quad  \end{align}\) (Using \(\text { a }^{2}-b^{2}=(a+b)(a-b)\)
)
\(\begin{align}   & \text{=}\frac{\text{si}{{\text{n}}^{\text{2}}}\text{ }\theta\text{ +co}{{\text{s}}^{\text{2}}}\text{ }\theta\text{ +2sin }\theta\text{ cos }\theta\text{ -1}}{\text{sin }\theta\text{ cos }\theta\text{ }} \end{align}\)
\(\begin{align}   & \text{=}\frac{\text{1+2sin }\theta\text{ cos }\theta\text{ -1}}{\text{sin }\theta\text{ cos}\theta\text{ }} \end{align}\)  (Using identify \(\sin ^{2} \theta+\cos ^{2} \theta=1\))
\(\begin{align}    \text{=}\frac{\text{2sin }\theta\text{ cos }\theta\text{ }}{\text{sin }\theta\text{ cos }\theta\text{ }}\text{= 2}\end{align}\)
Hence, option (C) is correct.

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

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),

\(\begin{align}  =\left( \frac{\text{1}}{\text{cosA}}\text{+}\frac{\text{sinA}}{\text{cosA}} \right)\text{(1}-\text{sinA)} \end{align}\)
\(\begin{align}  \text{=}\left( \frac{\text{1+sinA}}{\text{cosA}} \right)\text{(1}-\text{sinA)} \end{align}\) 
\(\begin{align} \text{=}\frac{\text{1}-\text{si}{{\text{n}}^{\text{2}}}\text{A}}{\text{cosA}}\end{align}\)
\(\begin{align} \text{=}\frac{\text{co}{{\text{s}}^{\text{2}}}\text{A}}{\text{cosA}}\quad \end{align}\) (By identify \(\sin ^{2} \theta+\cos ^{2} \theta=1\)
,Hence \(\rm 1- sin^2 θ = cos^2 θ\))
\(\begin{align} =\cos \text{A}  \\ \end{align}\)
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),

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

Hence, option (D) is correct

  
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