If -1-cos-tan -1-cos-x then sin x-

The correct option is A tan^2 (α/2 )Let cot^ 1√(cos α)=θ⇒√(cos α)=cot θ θ tan^ 1(cot θ)=x⇒θ tan^ 1(tan (π/2 θ ) )=x ⇒ 2 θ π/2=x ⇒ sin x = cos 2 θ = 1 1/co ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If -1√cosα-ta...

Question

If $c o t^{- 1} \sqrt{c o s α} - t a n^{- 1} \sqrt{c o s α} = x$ then sin x=

t a n^{2} (\frac{α}{2})

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c o t^{2} (\frac{α}{2})

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t a n α

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c o t (\frac{α}{2})

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Solution

The correct option is A $t a n^{2} (\frac{α}{2})$
Let $c o t^{- 1} \sqrt{c o s α} = θ \Rightarrow \sqrt{c o s α} = c o t θ θ - t a n^{- 1} (c o t θ) = x \Rightarrow θ - t a n^{- 1} (t a n (\frac{π}{2} - θ)) = x \Rightarrow 2 θ - \frac{π}{2} = x \Rightarrow s i n x = - c o s 2 θ = - \frac{1 - \frac{1}{c o s α}}{1 + \frac{1}{c o s α}} = t a n^{2} (\frac{α}{2})$

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If cot−1√cosα−tan−1√cosα=x then sin x=

The correct option is A tan2(α2)Let cot−1√cosα=θ⇒√cosα=cotθθ−tan−1(cotθ)=x⇒θ−tan−1(tan(π2−θ))=x⇒2θ−π2=x⇒sinx=−cos2θ=−1−1cosα1+1cosα=tan2(α2)

If $c o t^{- 1} \sqrt{c o s α} - t a n^{- 1} \sqrt{c o s α} = x$ then sin x=