Question

Prove that,
xcos (90-A) cos (90-A)+ tan (180°-A). cot (90+A)=
sinA.sin (180 - A)​

Answer 1

x = tan²A - 2sec²A

( check your question, if you havent got your answer )

Step-by-step explanation:

Given that

• x.cos(90+A) cos(90-A) + tan(180-A).cot(90-A) = sinAsin(180-A)

• => x.(-sinA).sinA + (-tanA).tanA = sinA.sinA

• => -xsin²A - tan²A = sin²A

• => sin²A + x sin²A = - tan²A

• => sin²A(1+x) = -tan²A

=> 1+x = $-tan²/sin²A = \frac{\frac{-sin^2A}{cos^2A} }{sin^2A}$

$= \frac{-sin^2A}{cos^2A*sin^2A}$

=> 1+x = -1/cos²A = -sec²A

=> x = -1-sec²A

=> x = -(sec²A-tan²A)-sec²A

=> x = tan²A-sec²A-sec²A

=> x = tan²A - 2sec²A

Answer 2

Step-by-step explanation:

=> 1+x = -1/cos²A = -sec²A

=> x = -1-sec²A

=> x = -(sec²A-tan²A)-sec²A

=> x = tan²A-sec²A-sec²A

=> x = tan²A - 2sec²A