Find
[cos²(45 + ∅) + cos²(45-∅) ] / [ tan( 60 +∅) .TAN (30 - ∅ ) + COSEC (75 + ∅) - SEC (15 - ∅) ]
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= 1
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⇒ {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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