Question

Evaluate: {cos²(45°+∅) + cos²(45°-∅) ÷ tan(60°+∅) tan(30°-∅)} +cosec(75°+∅) - sec(15-∅).
Do right ans

Answer 1

Answer:

1 is the correct answer

Step-by-step explanation:

⇒ {cos²(45°+θ) + cos²(45°-θ) ÷ tan(60°+θ)×tan(30°-θ)} + cosec(75°+θ) - sec(15°-θ)

First solving which is inside curly bracket.

⇒ [sin²{90°-(45°+θ)} + cos²(45°-θ) ÷ cot{90°-(60°+θ)}×tan(30°-θ)]

                                                     [∵ cosθ = sin(90°-θ)  &  tanθ = cot(90°-θ)]

⇒ [sin²(45°-θ) + cos²(45°-θ) ÷ cot(30°-θ)×tan(30°-θ)]

⇒ [1 ÷ 1]               [∵ sin²θ+cos²θ = 1  &  cotθ×tanθ = 1]

⇒ 1

Now solving the rest part.

⇒ 1 + cosec(75°+θ) - sec(15°-θ)

⇒ 1 + sec{90°-(75°+θ)} - sec(15°-θ)       [∵ cosecθ = sec(90°-θ)]

⇒ 1 + sec(15°-θ) - sec(15°-θ)

⇒ 1 + 0

⇒ 1     ←ANSWER

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