Question

Prove that sec square theta + cot square (90-theta) =2 cosec square (90-theta) -1 ???

Answer 1

HEYYA HERE IS YOUR ANSWER..........._______^^
TAKING L.H.S.
sec^2theetha + cot ^2(90-theetha)
sec^2theetha+ tan^2theetha
sec^2theetha + sec^2thhetha -1
2 sec^2theetha- 1....
NOW R.H.S
2cose^2(90-theetha)-1
2 sec^2theetha-1 ..
HENCE ,
LHS =RHS PROVED....

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HOPE THIS WILL HELP YOU...

Answer 2

$\sec^2 \theta+\cot^2 (90-\theta)=2\csc^2 (90-\theta) -1,$ proved.

Step-by-step explanation:

Prove that, $\sec^2 \theta+\cot^2 (90-\theta)=2\csc^2 (90-\theta) -1$.

L.H.S.$=\sec^2 \theta+\cot^2 (90-\theta)$

$=\sec^2 \theta+\tan^2 \theta$

Using the trigonometric identity,

$\tan \theta = \cot (90-\theta)$

$=\sec^2 \theta+\sec^2 \theta-1$

Using the trigonometric identity,

$\tan^2 \theta=\sec^2 \theta-1$

=$2\sec^2 \theta-1$

R.H.S.$=2\csc^2 (90-\theta) -1$

$=2\sec^2 \theta -1$

∴ L.H.S. = R.H.S.$=2\sec^2 \theta -1$

Hence, $\sec^2 \theta+\cot^2 (90-\theta)=2\csc^2 (90-\theta) -1$, proved.