Question

38. Prove that root (sec2 theta + cosec2 theta)
=tan theta plus cot theta

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Answer 1

Prove that:

$\sf \sqrt{sec^2\theta+cosec^2\theta}=tan\theta+cot\theta$

Proof:

L.H.S:

$\sf We\:know\:that,$

$\sf sec^2\theta=\dfrac{1}{cos^2\theta} \:and\: cosec^2\theta=\dfrac{1}{sin^2\theta}$

$\sf \therefore \sqrt{sec^2\theta+cosec^2\theta}=\sf\sqrt{ \dfrac{1}{cos^2\theta}+\dfrac{1}{sin^2\theta} }$

$\Rightarrow \sf\sqrt{ \dfrac{sin^2\theta+cos^2\theta}{cos^2\theta.sin^2\theta}$

$\Rightarrow \sf\sqrt{ \dfrac{1}{cos^2\theta.sin^2\theta}} \quad (as\: sin^2\theta+cos^2\theta=1)$

$\Rightarrow \sf \dfrac{1}{cos\theta.sin\theta}$

R.H.S:

$\sf tan\theta+cot\theta$

$\Rightarrow \sf \dfrac{sin\theta}{cos\theta}+\dfrac{cos\theta}{sin\theta}$

$\Rightarrow \sf \dfrac{sin^2\theta+cos^2\theta}{sin\theta.cos\theta}$

$\Rightarrow \sf \dfrac{1}{sin\theta.cos\theta}$

Hence,

L.H.S=R.H.S