Question

prove that sec square theta minus cot square 90 minus theta is equal to cos square 90 minus theta + cos square theta​

Answer 1

Answer:

the answer is above/////

Answer 2

$sec^2\ \theta -cot^2\ (90 - \theta) = cos^2(90 - \theta) + cos^2 \theta$

Solution:

Given that, we have to prove:

$sec^2\ \theta -cot^2\ (90 - \theta) = cos^2(90 - \theta) + cos^2 \theta$

Take the LHS

$sec^2\ \theta -cot^2\ (90 - \theta)$ -------- eqn 1

We know that,

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

Therefore, eqn 1 becomes

$sec^2 \theta - tan^2 \theta$ -------- eqn 2

We know that,

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

Thus, LHS = 1

Take the RHS

$cos^2(90 - \theta) + cos^2 \theta$ ----------- eqn 3

We know that,

$cos(90 - \theta) = sin\ \theta$

Therefore, eqn 3 becomes

$sin^2 \theta + cos^2 \theta\\\\We\ know\ that\\\\sin^2 \theta + cos^2 \theta = 1$

Thus, RHS = 1

Therefore,

LHS = RHS

Thus proved

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