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Question
Evaluate: 5 cosec2 45° – 3 sin2 90° + 5 cos 0°.
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Solution
5 cosec2 45° – 3 sin2 90° + 5 cos 0°
= `5(sqrt(2))^2 - 3(1)^2 + 5(1)`
= 10 – 3 + 5
= 12.
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Prove that: cot θ + tan θ = cosec θ·sec θ
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`
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= R.H.S.
∴ L.H.S. = R.H.S.
∴ cot θ + tan θ = cosec·sec θ
