Find the value of (i) sin⁻¹(cos(sin⁻¹dfrac√32)) (ii) cot(sin⁻¹dfrac35 + sin⁻¹dfrac45) (iii) tan(sin⁻¹dfrac35 +

Exercise 4.5 · Q3

Q. Find the value of

(i)

\sin^{-1}\left(\cos\left(\sin^{-1}\dfrac{\sqrt3}2\right)\right)

(ii)

\cot\left(\sin^{-1}\dfrac35 + \sin^{-1}\dfrac45\right)

(iii)

\tan\left(\sin^{-1}\dfrac35 + \cot^{-1}\dfrac32\right)

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Concept understanding — Composite and Sum/Difference Identities of Inverse Trigonometric Functions

These are the working-formula properties (Properties VI–X) for combining or composing inverse trig functions, together with the reference-triangle technique for a raw composite like $\tan(\sin^{-1}x)$ .

Composing a trig function with an unrelated inverse trig function (reference triangle). To evaluate $f(g^{-1}(x))$ where $f\ne g$ (e.g. $\cot(\sin^{-1}x)$ ), let $\theta=g^{-1}(x)$ , build a right triangle encoding $\theta$ from the definition of $g^{-1}$ (e.g. $\sin\theta=x$ gives opposite $=x$ , hypotenuse $=1$ , so adjacent $=\sqrt{1-x^2}$ by Pythagoras — taking the adjacent side non-negative since $\theta\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]$ keeps cosine $\ge0$ there), then read $f(\theta)$ straight off the triangle. This proves, e.g., $\tan(\sin^{-1}x)=\dfrac{x}{\sqrt{1-x^2}}$ , $-1<x<1$ .

Property VI (addition/subtraction formulas).

\sin^{-1}x+\sin^{-1}y=\sin^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right),\quad\text{if } x^2+y^2\le1 \text{ or } xy<0

\cos^{-1}x+\cos^{-1}y=\cos^{-1}\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right),\quad\text{if } x+y\ge0

\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\dfrac{x+y}{1-xy}\right),\quad\text{if } xy<1 \qquad \tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\dfrac{x-y}{1+xy}\right),\quad\text{if } xy>-1

with analogous subtraction forms for sine and cosine. Extending the tangent addition formula to three terms gives $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\left[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\right]$ ; setting the left side equal to $\pi$ and taking the tangent of both sides (which is $0$ ) proves the classical identity $x+y+z=xyz$ whenever $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi$ .

Property VII (double-angle-style formulas, from setting $y=x$ in Property VI).

2\tan^{-1}x=\tan^{-1}\left(\dfrac{2x}{1-x^2}\right),\ |x|<1 \qquad 2\tan^{-1}x=\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right),\ x\ge0 \qquad 2\tan^{-1}x=\sin^{-1}\left(\dfrac{2x}{1+x^2}\right),\ |x|\le1

Property VIII. $\sin^{-1}(2x\sqrt{1-x^2})=2\sin^{-1}x$ for $|x|\le\tfrac1{\sqrt2}$ , and $\sin^{-1}(2x\sqrt{1-x^2})=2\cos^{-1}x$ for $\tfrac1{\sqrt2}\le x\le1$ (the SAME left side splits into two different right sides depending on which half of $[-1,1]$ , i.e. which principal-range piece, $x$ falls in).

Property IX (mixed cofunction-and-Pythagorean substitutions). For $0\le x\le1$ : $\sin^{-1}x=\cos^{-1}\sqrt{1-x^2}$ and $\cos^{-1}x=\sin^{-1}\sqrt{1-x^2}$ ; for $-1\le x<0$ both pick up a sign/ ${}-\pi$ correction; and for $x>0$ , $\tan^{-1}x=\sin^{-1}\dfrac{x}{\sqrt{1+x^2}}=\cos^{-1}\dfrac1{\sqrt{1+x^2}}$ .

Property X (triple-angle-style formulas).

3\sin^{-1}x=\sin^{-1}(3x-4x^3),\ x\in\left[-\tfrac12,\tfrac12\right] \qquad 3\cos^{-1}x=\cos^{-1}(4x^3-3x),\ x\in\left[\tfrac12,1\right]

both proved the same way: set $x=\sin\theta$ (resp. $\cos\theta$ ) and substitute into the ordinary triple-angle trig identity.

Tip

The domain conditions attached to Properties VI–X are not decoration — they are exactly what keeps the sum of two principal-range angles from spilling OUTSIDE the target function's own principal range (which would otherwise force an extra $\pm\pi$ correction). Always check the condition before trusting the formula's plain output; e.g. proving $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi\Rightarrow x+y+z=xyz$ works by equating tangents (valid unconditionally, since $\tan\pi=0$ regardless of branch), which is why that particular proof needs no domain restriction even though the addition formula it is built from does.

Part (i) evaluates from the inside out; part (ii) spots that $\sin^{-1}\tfrac35+\sin^{-1}\tfrac45=\tfrac{\pi}2$ via the $3$ – $4$ – $5$ triangle; part (iii) uses the tangent addition formula on two reference-triangle angles.

Step 1. (i) Evaluate the innermost term. $\sin^{-1}\dfrac{\sqrt3}2=\dfrac{\pi}3$ .

Step 2. (i) Evaluate the cosine. $\cos\dfrac{\pi}3=\dfrac12$ .

Step 3. (i) Apply the outer $\sin^{-1}$ . $\sin^{-1}\left(\dfrac12\right)=\dfrac{\pi}6$ .

Step 4. (ii) Let $A=\sin^{-1}\dfrac35$ and $B=\sin^{-1}\dfrac45$ . Reference triangles: $\sin A=\dfrac35,\cos A=\dfrac45$ ; $\sin B=\dfrac45,\cos B=\dfrac35$ (both positive since $A,B\in\left[0,\dfrac{\pi}2\right]$ ).

Step 5. (ii) Compute $\sin(A+B)$ . $\sin A\cos B+\cos A\sin B=\dfrac35\cdot\dfrac35+\dfrac45\cdot\dfrac45=\dfrac9{25}+\dfrac{16}{25}=1$ .

Step 6. (ii) Conclude $A+B$ . Since $A,B\in\left[0,\dfrac{\pi}2\right]$ , $A+B\in[0,\pi]$ , and $\sin(A+B)=1\Rightarrow A+B=\dfrac{\pi}2$ .

Step 7. (ii) Evaluate. $\cot(A+B)=\cot\dfrac{\pi}2=0$ .

Step 8. (iii) Let $C=\sin^{-1}\dfrac35$ and $D=\cot^{-1}\dfrac32$ . Reference triangles: $\sin C=\dfrac35,\cos C=\dfrac45\Rightarrow\tan C=\dfrac34$ ; $\cot D=\dfrac32\Rightarrow\tan D=\dfrac23$ .

Step 9. (iii) Apply $\tan(C+D)=\dfrac{\tan C+\tan D}{1-\tan C\tan D}$ . $=\dfrac{\frac34+\frac23}{1-\frac34\cdot\frac23}=\dfrac{\frac9{12}+\frac8{12}}{1-\frac12}=\dfrac{\frac{17}{12}}{\frac12}=\dfrac{17}6$ .

✓Final answer

\dfrac{\pi}6

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

CBSE 2020Set ANNUAL1 markMCQ
$\tan^{-1}\left(\dfrac{1}{4}\right) + \tan^{-1}\left(\dfrac{2}{9}\right)$
(a) $\tan^{-1}\left(\dfrac{1}{2}\right)$
(b) $\dfrac{1}{2}\cos^{-1}\left(\dfrac{3}{5}\right)$
(c) $\dfrac{1}{2}\sin^{-1}\left(\dfrac{3}{5}\right)$
(d) $\dfrac{1}{2}\tan^{-1}\left(\dfrac{3}{5}\right)$
›Reveal solutionSolution
Combining the two inverse tangents gives $\tan^{-1}(1/2)$ , which by the half-angle tangent formula equals $\dfrac12\cos^{-1}\!\left(\dfrac35\right)$ .

Use the addition formula $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\!\left(\dfrac{x+y}{1-xy}\right)$ , valid here since $x=\dfrac14,\ y=\dfrac29$ give $xy=\dfrac{2}{36}=\dfrac{1}{18}<1$ .

Compute $x+y=\dfrac14+\dfrac29=\dfrac{9}{36}+\dfrac{8}{36}=\dfrac{17}{36}$ .

Compute $1-xy=1-\dfrac{1}{18}=\dfrac{17}{18}$ .

So $\dfrac{x+y}{1-xy}=\dfrac{17/36}{17/18}=\dfrac{17}{36}\times\dfrac{18}{17}=\dfrac{18}{36}=\dfrac12$ .

Hence $\tan^{-1}\!\left(\dfrac14\right)+\tan^{-1}\!\left(\dfrac29\right)=\tan^{-1}\!\left(\dfrac12\right)$ .

Now express this using $\cos^{-1}(3/5)$ : let $\theta=\cos^{-1}\!\left(\dfrac35\right)$ , so $\cos\theta=\dfrac35$ and, since $\theta$ is acute, $\sin\theta=\sqrt{1-\left(\frac35\right)^2}=\dfrac45$ .

Use the half-angle formula $\tan\dfrac{\theta}{2}=\dfrac{\sin\theta}{1+\cos\theta}=\dfrac{4/5}{1+3/5}=\dfrac{4/5}{8/5}=\dfrac12$ .

So $\dfrac{\theta}{2}=\tan^{-1}\!\left(\dfrac12\right)$ , i.e. $\tan^{-1}\!\left(\dfrac12\right)=\dfrac12\cos^{-1}\!\left(\dfrac35\right)$ .

Combining steps 5 and 8: $\tan^{-1}\!\left(\dfrac14\right)+\tan^{-1}\!\left(\dfrac29\right)=\dfrac12\cos^{-1}\!\left(\dfrac35\right)$ , matching the standard textbook identity.

✓Final answer
$\tan^{-1}\!\left(\dfrac14\right)+\tan^{-1}\!\left(\dfrac29\right)=\dfrac12\cos^{-1}\!\left(\dfrac35\right)$

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