1.

Show that:cos²(45° + ø) + cos²(45° - ø)/tan(60° + ø) tan(30° - ø)= 1​

Answer»

SOLUTION:

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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