Question

prove- under root sec2 THETA + COSEC2 THETA = TAN THETA + COT THETA

Answer 1

To prove:

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

Proof:

$\sf{L.H.S.=\sqrt{sec^{2}\theta+cosec^{2}\theta}}$

$\sf\blue{sec\theta=\dfrac{1}{cos\theta} \ and \ cosec\theta=\dfrac{1}{sin\theta}}$

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

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

$\sf\blue{sin^{2}\theta+cos^{2}\theta=1}$

$\sf{=\sqrt{\dfrac{1}{(sin\theta.cos\theta)^{2}}}}$

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

$\sf{R.H.S.=tan\theta+cot\theta}$

$\sf\blue{tan\theta=\dfrac{sin\theta}{cos\theta} \ and \ cot\theta=\dfrac{cos\theta}{sin\theta}}$

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

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

$\sf\blue{sin^{2}\theta+cos^{2}\theta=1}$

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

$\sf{=L.H.S.}$

$\sf\purple{\tt{Hence, \ proved.}}$

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