Math, asked by neelamyadav11081982, 10 months ago

Evaluate {cos²(45°-Q)+cos²(45°-Q)/tan(60°+Q)tan(30°-Q)}+ cosec(75-Q)-sec(15-Q)

Answers

Answered by dududakeshi
0

Answer:

Step-by-step explanation:LHS = cos ²(45°+ϴ) + cos²(45° - ϴ) / tan(60° +ϴ) tan(30 - ϴ) = 1

= sin²[90° - (45°+ϴ)]+ cos²(45° - ϴ) / cot [90° - (60° +ϴ)] tan(30 - ϴ)

[ Cosϴ= sin(90°-ϴ) & cot ϴ= tan (90°-ϴ)]

= sin²(45°-ϴ)+ cos²(45° - ϴ) / cot 30° - ϴ)] tan(30° - ϴ)

= 1/1 = RHS

[ sin²ϴ + cos²ϴ= 1 and tanϴcotϴ=1]

HOPE THIS WILL HELP YOU...

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