Question

please answer the 40th q

Answer 1

$\large{\underline{\underline{\mathfrak{\green{\sf{QUESTION}}}}}}$.

To Proof,

$\pink{\sf{\:\left(1+\dfrac{1}{\tan^2 \theta}\right)\times \left(1+\dfrac{1}{\cot^2 \theta}\right)\:=\dfrac{1}{(\sin^2 \theta - \sin^4 \theta)}}}$

$\large{\underline{\underline{\mathfrak{\green{\sf{Prove}}}}}}$.

L.H.S.

$\pink{\sf{\:\left(1+\dfrac{1}{\tan^2 \theta}\right)\times \left(1+\dfrac{1}{\cot^2 \theta}\right)}}$

$\:\:\:\:\:\:\:\:\star\green{\sf{\:\tan \theta\:=\dfrac{\sin \theta}{\cos \theta}}}$

$\:\:\:\:\:\:\:\:\star\green{\sf{\:\cot \theta\:=\dfrac{\cos \theta}{\sin \theta}}}$

$\mapsto\sf{\:\left(1+\dfrac{1}{\dfrac{\sin^2 \theta}{\cos^2 \theta}}\right)\times \left(1+\dfrac{1}{\dfrac{\cos^2 \theta}{\sin^2 \theta}}\right)}$

$\mapsto\sf{\:\left(1+\dfrac{\cos^2 \theta}{\cos^2 \theta}\right)\times \left(1+\dfrac{\sin^2 \theta}{\cos^2 \theta}\right)}$

$\mapsto\sf{\:\left(\dfrac{\sin^2 \theta+\cos^2 \theta}{\sin^2 \theta}\right)\times \left(\dfrac{\cos^2 \theta+\sin^2 \theta}{\cos^2 \theta}\right)}$

$\:\:\:\:\:\:\:\:\:\star\green{\sf{\:(\sin^2 \theta+\cos^2 \theta\:=\:1)}}$

$\mapsto\sf{\:\left(\dfrac{1}{\sin^2 \theta}\right)\times \left(\dfrac{1}{\cos^2 \theta}\right)}$

$\:\:\:\:\:\:\:\:\star\green{\sf{\:\left(\cos^2 \theta\:=\:1-\sin^2 \theta\right)}}$

$\mapsto\sf{\:\left(\dfrac{1}{\sin^2 \theta}\right)\times \left(\dfrac{1}{(1-\sin^2 \theta}\right)}$

$\mapsto\pink{\sf{\:\left(\dfrac{1}{\sin^2 \theta - \sin^4 \theta}\right)}}$

= R.H.S.

That's Proved,

_____________________

Answer 2

Answer:

refer to the above attachment for the solution.