The value of the expression $\tan\left(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt5}\right)$ is [Hint: $\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$] (A) $2+\sqrt5$ (B) $\sqrt5-2$ (C) $\frac{\sqrt5+2}{2}$ (D) $\sqrt5+2$

> [!TLDR] > The key idea is to let $\cos^{-1}\frac{2}{\sqrt5} = \theta$, then use the half-angle formula for tangent. The value simplifies to $\sqrt5 - 2$, which matches option (B). We start with the expression $\tan\left(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt5}\right)$. The hint gives us a direct path: if we set $\theta = \cos^{-1}\frac{2}{\sqrt5}$, then $\cos\theta = \frac{2}{\sqrt5}$, and we need $\tan\frac{\theta}{2}$. The formula $\tan\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$ is perfect — but we must be careful about the sign. Since $\theta = \cos^{-1}\frac{2}{\sqrt5}$ lies in $[0, \pi]$, and $\frac{2}{\sqrt5} \approx 0.894$ is positive, $\theta$ is in the first quadrant ($0 0$. We take the positive square root. 1. **Set up the substitution** Let $\theta = \cos^{-1}\frac{2}{\sqrt5}$. Then $\cos\theta = \frac{2}{\sqrt5}$. 2. **Apply the half-angle formula** $$ \tan\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} $$ Substitute $\cos\theta$: $$ \tan\frac{\theta}{2} = \sqrt{\frac{1 - \frac{2}{\sqrt5}}{1 + \frac{2}{\sqrt5}}} $$ 3. **Simplify the fraction inside the square root** Multiply numerator and denominator by $\sqrt5$ to clear the fraction: $$ \frac{1 - \frac{2}{\sqrt5}}{1 + \frac{2}{\sqrt5}} = \frac{\sqrt5 - 2}{\sqrt5 + 2} $$ So we have: $$ \tan\frac{\theta}{2} = \sqrt{\frac{\sqrt5 - 2}{\sqrt5 + 2}} $$ 4. **Rationalise the denominator inside the square root** …

The value of the expression tan((1)/(2)cos⁻¹(2)/(√5)) is [Hint: tan(θ)/(2)=√((1-cosθ)/(1+cosθ))] (A) 2+√5 (B)

NCERT Exemplar · Q33

\tan\left(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt5}\right)

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Appeared in past exams:KCET 2018· Set A-1· 1mexact

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Concept understanding — Trigonometric Simplification

Trigonometric Simplification

You know $\sin^2 x + \cos^2 x = 1$ — but the skill of turning a messy trig expression into that kind of clean form is trigonometric simplification. Because sines, cosines and their relatives are all tied together by identities from the unit circle, a tangled combination can almost always be rewritten as something shorter: a single term, a constant, or an easier combination.

Your core toolkit

\sin^2\theta+\cos^2\theta=1,\quad 1+\tan^2\theta=\sec^2\theta,\quad 1+\cot^2\theta=\csc^2\theta

How the process feels

Simplify $\dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x}$ . Over a common denominator the numerator is $\sin^2 x+(1+\cos x)^2 = \sin^2 x + 1 + 2\cos x + \cos^2 x$ . The Pythagorean identity turns $\sin^2 x+\cos^2 x$ into $1$ , giving $2+2\cos x = 2(1+\cos x)$ , so

\frac{2(1+\cos x)}{\sin x\,(1+\cos x)} = \frac{2}{\sin x} = 2\csc x.

A two-term sum collapses to one term.

Strategies that usually work

Convert everything to sines and cosines — cancellations then appear.
Spot Pythagorean pairs and replace them with $1$ (or $\sec^2$ , $\csc^2$ ).
Factor and cancel as you would with ordinary algebra.
Multiply by a conjugate — e.g. multiply $\frac{1}{1+\sin x}$ by $\frac{1-\sin x}{1-\sin x}$ to unlock a Pythagorean identity. …

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The key idea is to let $\cos^{-1}\frac{2}{\sqrt5} = \theta$ , then use the half-angle formula for tangent. The value simplifies to $\sqrt5 - 2$ , which matches option (B).

We start with the expression $\tan\left(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt5}\right)$ . The hint gives us a direct path: if we set $\theta = \cos^{-1}\frac{2}{\sqrt5}$ , then $\cos\theta = \frac{2}{\sqrt5}$ , and we need $\tan\frac{\theta}{2}$ . The formula $\tan\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$ is perfect — but we must be careful about the sign. Since $\theta = \cos^{-1}\frac{2}{\sqrt5}$ lies in $[0, \pi]$ , and $\frac{2}{\sqrt5} \approx 0.894$ is positive, $\theta$ is in the first quadrant ( $0 < \theta < \frac{\pi}{2}$ ). Then $\frac{\theta}{2}$ is also in the first quadrant, so $\tan\frac{\theta}{2} > 0$ . We take the positive square root.

Set up the substitution

Let $\theta = \cos^{-1}\frac{2}{\sqrt5}$ . Then $\cos\theta = \frac{2}{\sqrt5}$ .
Apply the half-angle formula

\tan\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}

Substitute $\cos\theta$ :

\tan\frac{\theta}{2} = \sqrt{\frac{1 - \frac{2}{\sqrt5}}{1 + \frac{2}{\sqrt5}}}

Simplify the fraction inside the square root Multiply numerator and denominator by $\sqrt5$ to clear the fraction:

\frac{1 - \frac{2}{\sqrt5}}{1 + \frac{2}{\sqrt5}} = \frac{\sqrt5 - 2}{\sqrt5 + 2}

So we have:

\tan\frac{\theta}{2} = \sqrt{\frac{\sqrt5 - 2}{\sqrt5 + 2}}

Rationalise the denominator inside the square root …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Higher Secondary (+2 Stage) Examination 2025Set ANNUAL3 marks
$\tan^{-1}\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x$
›Reveal solutionSolution

The substitution $x=\cos2\theta$ turns $\sqrt{1\pm x}$ into a clean $\sqrt2\cos\theta$ / $\sqrt2\sin\theta$ pair, collapsing the whole expression to $\tan(\pi/4-\theta)$ .

Let $x=\cos2\theta$ , so $\theta=\dfrac12\cos^{-1}x$ . Using $1+\cos2\theta=2\cos^2\theta$ and $1-\cos2\theta=2\sin^2\theta$ :

$\sqrt{1+x}=\sqrt{2\cos^2\theta}=\sqrt2\,\cos\theta,\qquad \sqrt{1-x}=\sqrt{2\sin^2\theta}=\sqrt2\,\sin\theta$

(taking $\theta$ in a range where $\cos\theta,\sin\theta\ge0$ , consistent with the given domain $-\tfrac{1}{\sqrt2}\le x\le1$ ).

Substitute into the given expression:

$\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\dfrac{\sqrt2\cos\theta-\sqrt2\sin\theta}{\sqrt2\cos\theta+\sqrt2\sin\theta}=\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}.$

Divide numerator and denominator by $\cos\theta$ :

$=\dfrac{1-\tan\theta}{1+\tan\theta}=\tan\left(\dfrac{\pi}{4}-\theta\right),$ …

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Q.The value of the expression $\tan\left(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt5}\right)$ is [Hint: $\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$ ]
(A) $2+\sqrt5$
(B) $\sqrt5-2$
(C) $\frac{\sqrt5+2}{2}$
(D) $\sqrt5+2$

Concept understanding — Trigonometric Simplification

Trigonometric Simplification

Your core toolkit

How the process feels

Strategies that usually work

Method: $\tan$ of half an inverse-cosine angle

Steps

Common Mistakes

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