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The expression $\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ}$ is equal to

व्यंजक $\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ}$ का मान है

tanθ + cotθ + 1

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tanθ + cotθ – 1

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(1 + sinθ·cosθ)(secθ·cosecθ)

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(1 – sinθ·cosθ)(secθ·cosecθ)

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Solution

The correct option is C (1 + sinθ·cosθ)(secθ·cosecθ)
Solution

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Similar questions

cos θ = \frac{1}{3}, (θ \in R)

cos θ = \frac{1}{3}, (θ \in R)

Is the identity $\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ} = 1 + tan θ + cot θ$ true or false?

\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ} = 1 + tan θ + cot θ = 1 + sec θ csc θ

\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ} = 1 + tan θ + cot θ

\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = 1 + t a n θ + c o t θ = s e c θ c o s e c θ + 1

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