Question

evaluate sec square(90-theta)-cot square theta/2(sin square 25+sin square 65)

Answer 1

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Answer 2

Answer:

$\frac{\sec^2(90-\theta)-\cot^2\theta}{2(\sin^2 25+\sin^2 65)}=\frac{1}{2}$    

Step-by-step explanation:

Given : Expression $\frac{\sec^2(90-\theta)-\cot^2\theta}{2(\sin^2 25+\sin^2 65)}$

To find : Evaluate the expression ?

Solution :

Expression $\frac{\sec^2(90-\theta)-\cot^2\theta}{2(\sin^2 25+\sin^2 65)}$

$=\frac{\csc^2\theta-\cot^2\theta}{2(\sin^2 (90-65)+\sin^2 65)}$

$=\frac{1}{2(\cos^2 65+\sin^2 65)}$

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

$=\frac{1}{2(1)}$

$=\frac{1}{2}$

Therefore, $\frac{\sec^2(90-\theta)-\cot^2\theta}{2(\sin^2 25+\sin^2 65)}=\frac{1}{2}$