Exercise-8.3, Class 10th, Maths, Exercise 8, NCERT

Q 1. Express sin θ, sec θ and tan θ in terms of cot θ.

Let $\cot \theta =c$

$$\tan \theta =\frac{1}{c},1+{\cot }^2\theta ={\csc }^2\theta \mathrm{.}$$

So,

$$\csc \theta =\sqrt{1+c^2}\Rightarrow \sin \theta =\frac{1}{\sqrt{1+c^2}}\mathrm{.}$$

Also,

$$1+{\tan }^2\theta ={\sec }^2\theta \Rightarrow \sec \theta =\sqrt{1+{\tan }^2\theta }=\sqrt{1+\frac{1}{c^2}}=\frac{\sqrt{1+c^2}}{c}\mathrm{.}$$
$$\boxed{{\displaystyle \sin \theta =\frac{1}{\sqrt{1+{\cot }^2\theta }},\tan \theta =\frac{1}{\cot \theta },\sec \theta =\frac{\sqrt{1+{\cot }^2\theta }}{\cot \theta }\mathrm{.}}}$$

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Q 2. Express all trigonometric ratios of θ in terms of sec θ.

Let $\sec \theta =s\mathrm{}$

Then $\cos \theta =\frac{1}{s}\mathrm{}$

$$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{1}{s^2}}=\frac{\sqrt{s^2-1}}{s}\mathrm{}$$

Then:

$$\tan \theta =\frac{\sin \theta }{\cos \theta }=\sqrt{s^2-1},\cot \theta =\frac{1}{\tan \theta }=\frac{1}{\sqrt{s^2-1}},\csc \theta =\frac{1}{\sin \theta }=\frac{s}{\sqrt{s^2-1}}\mathrm{.}$$
$$\boxed{{\displaystyle \begin{array}{rl}{\displaystyle \cos \theta }& {\displaystyle =\frac{1}{\sec \theta },}\\ {\displaystyle \sin \theta }& {\displaystyle =\frac{\sqrt{{\sec }^2\theta -1}}{\sec \theta },}\\ {\displaystyle \tan \theta }& {\displaystyle =\sqrt{{\sec }^2\theta -1},}\\ {\displaystyle \cot \theta }& {\displaystyle =\frac{1}{\sqrt{{\sec }^2\theta -1}},}\\ {\displaystyle \csc \theta }& {\displaystyle =\frac{\sec \theta }{\sqrt{{\sec }^2\theta -1}}\mathrm{.}}\end{array}}}$$

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Q 3. Evaluate each expression (choose correct option).

(i) $9{\sec }^2\theta -9{\tan }^2\theta$

$$9({\sec }^2\theta -{\tan }^2\theta )=9(1)=\boxed{{\displaystyle 9.}}$$

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(ii) $(1+\tan \theta +\sec \theta )(1+\cot \theta -\csc \theta )$

$$=\frac{\cos \theta +\sin \theta +1}{\cos \theta }\times \frac{\sin \theta +\cos \theta -1}{\sin \theta }=\frac{(\sin \theta +\cos \theta {)}^2-1}{\sin \theta \cos \theta }\mathrm{.}$$
$$(\sin \theta +\cos \theta {)}^2-1=1+2\sin \theta \cos \theta -1=2\sin \theta \cos \theta \mathrm{.}$$
$$\frac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }=\boxed{{\displaystyle 2.}}$$

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(iii) $(\sec \theta +\tan \theta )(1-\sin \theta )$

$$=\frac{1+\sin \theta }{\cos \theta }(1-\sin \theta )=\frac{1-{\sin }^2\theta }{\cos \theta }=\frac{{\cos }^2\theta }{\cos \theta }=\boxed{{\displaystyle \cos \theta \mathrm{.}}}$$

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(iv) ${\displaystyle \frac{1+{\tan }^2\theta }{1+{\cot }^2\theta }}$

$$=\frac{{\sec }^2\theta }{{\csc }^2\theta }=\frac{{\sin }^2\theta }{{\cos }^2\theta }=\boxed{{\displaystyle {\tan }^2\theta \mathrm{.}}}$$

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Q 4. Prove the following identities.

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(i) $(\csc \theta -\cot \theta {)}^2={\displaystyle \frac{1-\cos \theta }{1+\cos \theta }}$

$$\csc \theta -\cot \theta =\frac{1-\cos \theta }{\sin \theta },(\csc \theta -\cot \theta {)}^2=\frac{(1-\cos \theta {)}^2}{{\sin }^2\theta }\mathrm{.}$$$${\sin }^2\theta =(1-\cos \theta )(1+\cos \theta )\Rightarrow \frac{(1-\cos \theta {)}^2}{(1-\cos \theta )(1+\cos \theta )}=\frac{1-\cos \theta }{1+\cos \theta }\mathrm{.}$$
$$\boxed{{\displaystyle LHS=RHS\mathrm{.}}}$$

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(ii) $(\csc \theta +\cot \theta {)}^2={\displaystyle \frac{1+\cos \theta }{1-\cos \theta }}$

$$\csc \theta +\cot \theta =\frac{1+\cos \theta }{\sin \theta }\mathrm{.}$$$$(\csc \theta +\cot \theta {)}^2=\frac{(1+\cos \theta {)}^2}{{\sin }^2\theta }=\frac{(1+\cos \theta {)}^2}{(1-\cos \theta )(1+\cos \theta )}=\frac{1+\cos \theta }{1-\cos \theta }\mathrm{.}$$
$$\boxed{{\displaystyle LHS=RHS\mathrm{.}}}$$

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(iii) $(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$

$$(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )={\sec }^2\theta -{\tan }^2\theta =1.$$

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(iv) $(\sec \theta +\tan \theta )(1-\sin \theta )=\cos \theta$

$$\sec \theta +\tan \theta =\frac{1+\sin \theta }{\cos \theta }\mathrm{}$$

$$(\sec \theta +\tan \theta )(1-\sin \theta )=\frac{1-{\sin }^2\theta }{\cos \theta }=\frac{{\cos }^2\theta }{\cos \theta }=\cos \theta \mathrm{.}$$

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(v) ${\displaystyle \frac{1+\sin \theta }{\cos \theta }}=\sec \theta +\tan \theta$

$$\frac{1+\sin \theta }{\cos \theta }=\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=\sec \theta +\tan \theta \mathrm{.}$$

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(vi) ${\displaystyle \frac{1-\sin \theta }{\cos \theta }}=\sec \theta -\tan \theta$

$$\frac{1-\sin \theta }{\cos \theta }=\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }=\sec \theta -\tan \theta \mathrm{.}$$

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(vii) ${\displaystyle \frac{1+\tan \theta }{1+\cot \theta }}=\tan \theta$

$$\frac{1+\tan \theta }{1+\cot \theta }=\frac{1+\tan \theta }{1+\frac{1}{\tan \theta }}=\frac{1+\tan \theta }{\frac{1+\tan \theta }{\tan \theta }}=\tan \theta \mathrm{.}$$

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(viii) ${\displaystyle \frac{1+\sin \theta }{1-\sin \theta }}=(\sec \theta +\tan \theta {)}^2$

RHS:

$$(\sec \theta +\tan \theta {)}^2={\left(\frac{1+\sin \theta }{\cos \theta }\right)}^2=\frac{(1+\sin \theta {)}^2}{{\cos }^2\theta }=\frac{(1+\sin \theta {)}^2}{(1-\sin \theta )(1+\sin \theta )}=\frac{1+\sin \theta }{1-\sin \theta }\mathrm{.}$$
$$\boxed{{\displaystyle LHS=RHS\mathrm{.}}}$$

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(ix) $\sin \theta +\cos \theta =\sqrt{2}\sin (\theta +45\mathrm{°})$

$$\sin (\theta +45\mathrm{°})=\sin \theta \cos 45\mathrm{°}+\cos \theta \sin 45\mathrm{°}=\frac{\sin \theta +\cos \theta }{\sqrt{2}}\mathrm{.}$$

Multiply by √2:

$$\sin \theta +\cos \theta =\sqrt{2}\sin (\theta +45\mathrm{°})\mathrm{.}$$

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(x) ${\sin }^4\theta +{\cos }^4\theta =1-\frac{1}{2}{\sin }^{2}2\theta$

$${\sin }^4\theta +{\cos }^4\theta =({\sin }^2\theta +{\cos }^2\theta {)}^2-2{\sin }^2\theta {\cos }^2\theta =1-2{\sin }^2\theta {\cos }^2\theta \mathrm{.}$$

But $\sin 2\theta =2\sin \theta \cos \theta \Rightarrow {\sin }^{2}2\theta =4{\sin }^2\theta {\cos }^2\theta \mathrm{.}$$$2{\sin }^2\theta {\cos }^2\theta =\frac{1}{2}{\sin }^{2}2\theta \mathrm{.}$$$$\therefore {\sin }^4\theta +{\cos }^4\theta =1-\frac{1}{2}{\sin }^{2}2\theta \mathrm{.}$$