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प्रश्न
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
विकल्प
0
1
-1
2
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उत्तर
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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संबंधित प्रश्न
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Choose the correct alternative:
sec2θ – tan2θ =?
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
