Question

Cos2(45+a)+cos2(45-a)/tan(60+a)tan(30-a)+(cot30+sin90)(tan60-sec0)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

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

[ sin²ϴ + cos²ϴ= 1 and tanϴcotϴ=1]

HOPE THIS WILL HELP YOU...