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Question
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Options
0
1
-1
2
Solution
The given expression is `cot θ/(cotθ-cot 3θ)+tanθ/(tanθ-tan3θ)`
Simplifying the given expression, we have
`cotθ/(cotθ-cot3θ)+ tanθ/(tanθ-tan3θ)`
= `(cosθ/sin)/(cosθ/sinθ-(cos3θ)/(sin3θ))+(sinθ/cosθ)/(sinθ/sinθ-(sin3θ)/(cos3θ))`
=` (cosθ/sinθ)/((cosθsin 3θ-cos3θsinθ)/(sinθ sin3θ))+ (sin θ/cos θ)/((sinθ cos3θ-sin3θ cosθ)/(cosθ cos3θ))`
=` (cosθ sin3θ)/(cosθ sin3θ-cos3θsinθ)+(sinθ cos3θ)/(sinθ cos3θ-sin3θ sinθ)`
=`(cosθ sin3θ)/(cosθsinθ-cos3θsinθ)+(cos3θ sinθ)/(-1(cosθ sin3θ-cos3θ sinθ))`
`= (cosθ sin3θ)/(cosθ sin3θ-cos3θsinθ)-(cos3θsinθ)/(cosθsin3θ-cos3θsinθ)`
`=(cosθsin3θ-cos3θsinθ)/(cosθsin3θ-cos3θsinθ)`
=1
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
