Question

plz slove
urgent hai
I will mark you
as brainliest.


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Answer 1

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

Prove That

$\mapsto\pink{\sf{\dfrac{(1+\sec \theta - \tan \theta)}{(1+\sec \theta + \tan \theta)}\:=\:\left(\dfrac{1-\sin \theta}{\cos \theta}\right)}}$

$\large\pink{\underline{\underline{\mathfrak{\green{\sf{PROVE}}}}}}$.

L.H.S.

$\mapsto\pink{\sf{\dfrac{(1+\sec \theta - \tan \theta)}{(1+\sec \theta + \tan \theta)}}}$

$\:\:\:\:\:\green{\sf{\:(\sec^2 \theta - \tan^2 \theta\:=\:1)}}$

$\mapsto\sf{\dfrac{(\sec^2 \theta - \tan^2 \theta)+(\sec \theta - \tan \theta}{(1+\sec \theta + \tan \theta)}}$

$\:\:\:\:\:\green{\sf{\:(a^2-b^2)\:=\:(a+b)(a-b)}}$

$\mapsto\sf{\dfrac{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)+(\sec \theta - \tan \theta)}{(1+\sec \theta + \tan \theta)}}$

$\:\:\:\:\:\green{\sf{\:(take\:common)}}$

$\mapsto\sf{\dfrac{(\sec \theta - \tan \theta)\cancel{(\sec \theta + \tan \theta + 1)}}{\cancel{(1+\sec \theta + \tan \theta )}}}$

$\mapsto\sf{\:(\sec \theta - \tan \theta)}$

$\:\:\:\:\:\green{\sf{\:\left(\sec \theta \:=\dfrac{1}{\cos \theta}\:\:and\:\tan \theta \:=\dfrac{\sin \theta}{\cos \theta}\right)}}$

$\mapsto\sf{\dfrac{1}{\cos \theta}\:-\dfrac{\sin \theta}{\cos \theta}}$

$\mapsto\pink{\sf{\left(\dfrac{1-\sin \theta}{\cos \theta}\right)}}$

$\:\:\:\:\red{\sf{\:=RHS}}$

Proved

________________,_,____

Answer 2

Answer:

your answer attached in the photo