InterviewSolution
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InterviewSolution
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2 cos*(45°+0)+ cos®(45° - 9I 60. Evaluate: “tan (60°7 6 tan (30°=8) +cosec (75° + 8) - sec (15° - 0). [CBSE 2011] | |
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Answer» 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] |
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