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Question
Evaluate 2 sec2 θ + 3 cosec2 θ – 2 sin θ cos θ if θ = 45°.
Solution
2 sec2 θ + 3 cosec2 θ – 2 sin θ cos θ
= `2(sqrt(2))^2 + 3(sqrt(2))^2 - 2. 1/sqrt(2). 1/sqrt(2)` ...[∵ θ = 45°]
= `2 xx 2 + 3 xx 2 - 2 xx 1/2`
= 4 + 6 – 1
= 9
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Proof: L.H.S. = cot θ + tan θ
= `square/square + square/square` ......`[∵ cot θ = square/square, tan θ = square/square]`
= `(square + square)/(square xx square)` .....`[∵ square + square = 1]`
= `1/(square xx square)`
= `1/square xx 1/square`
= cosec θ·sec θ ......`[∵ "cosec" θ = 1/square, sec θ = 1/square]`
= R.H.S.
∴ L.H.S. = R.H.S.
∴ cot θ + tan θ = cosec·sec θ
