Question

Evaluate [sin square theta + sin square (90° - theta)] / [3(sec square 61° - cot square 29°)] - [3 cot square 30° sin square 54° sec square 36°] / [2(cosec square 65° - tan square 25°)] .

Answer 1

Answer: The value of the given expression is $-\frac{25}{6}$.

Explanation:

The given expression is,

$\frac{\sin^2\theta+\sin^2(90-\theta)}{3[\sec^2(61)-\cot^2(29)]}-\frac{3\cot^2(30)\sin^2(54)\sec^2(36)}{2[\csc^2(65)-\tan^2(25)]}$

$\frac{\sin^2\theta+\sin^2(90-\theta)}{3[\sec^2(61)-\cot^2(90-61)]}-\frac{3\cot^2(30)\sin^2(54)\sec^2(90-54)}{2[\csc^2(65)-\tan^2(90-65)]}$

Using quadrant concepts we get,

$\frac{\sin^2\theta+\cos^2(\theta)}{3[\sec^2(61)-\tan^2(61)]}-\frac{3\cot^2(30)\sin^2(54)\csc^2(54)}{2[\csc^2(65)-\cot^2(65)]}$

Using the formulas,

$\sin^2\theta+\cos^2(\theta)=1\\\sec^2(\theta)-\tan^2(\theta)=1\\\csc^2(\theta)-\cot^2(\theta)=1\\\csc\theta=\frac{1}{sin^2\theta}$

We get,

$\frac{1}{3(1)}-\frac{3\cot^2(30)(1)}{2(1)}$

$cot(30)=\sqrt{3}$

$\frac{1}{3} -\frac{3\times 3\times 1}{2\times 1}$

$\frac{2-27}{3(2)}$

$\frac{-25}{6}$

Therefore, the value of the expression is $-\frac{25}{6}$.

Answer 2

Answer:

Step-by-step explanation: