Prove that $\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}$.

> [!TLDR] > We prove the identity by converting the sum of two inverse sines into an inverse sine using the sine addition formula and checking the angle lies in the correct range. The result is $\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}$. The core idea here is that adding two angles and then taking the sine of their sum gives us a way to combine the inverse sine terms. But we must be careful: inverse sine returns an angle in $[-\pi/2, \pi/2]$, so we need to verify that the sum of our two angles also falls in that range — otherwise the inverse sine on the right might give a different angle (like its supplement). This is a classic pitfall in inverse trigonometric identities. Let’s denote: $$ A = \sin^{-1}\frac{8}{17}, \quad B = \sin^{-1}\frac{3}{5} $$ We want to show $A+B = \sin^{-1}\frac{77}{85}$. 1. **Find $\cos A$ and $\cos B$.** Since $\sin A = \frac{8}{17}$ and $A$ is in $[-\pi/2, \pi/2]$, $\cos A \ge 0$. Using $\sin^2 A + \cos^2 A = 1$: $$ \cos A = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17} $$ Similarly, $\sin B = \frac{3}{5}$, so: $$ \cos B = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} $$ 2. **Apply the sine addition formula.** $$ \sin(A+B) = \sin A \cos B + \cos A \sin B = \frac{8}{17} \cdot \frac{4}{5} + \frac{15}{17} \cdot \frac{3}{5} $$ $$ = \frac{32}{85} + \frac{45}{85} = \frac{77}{85} $$ So $\sin(A+B) = \frac{77}{85}$. This tells us that $A+B$ is some angle whose sine is $\frac{77}{85}$. But it could be either $\sin^{-1}\frac{77}{85}$ or $\pi - \sin^{-1}\frac{77}{85}$, since sine is positive in both first and second quadrants. 3. **Check the range of $A+B$.** We need to see whether $A+B$ lies in $[-\pi/2, \pi/2]$ (the principal range of $\sin^{-1}$). - $A = \sin^{-1}\frac{8}{17}$. Since $\frac{8}{17} \approx 0.4706$, $A$ is about $0.49$ radians ($28

Prove that sin⁻¹(8)/(17)+sin⁻¹(3)/(5)=sin⁻¹(77)/(85).

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\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}

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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). …

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We prove the identity by converting the sum of two inverse sines into an inverse sine using the sine addition formula and checking the angle lies in the correct range. The result is $\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}$ .

The core idea here is that adding two angles and then taking the sine of their sum gives us a way to combine the inverse sine terms. But we must be careful: inverse sine returns an angle in $[-\pi/2, \pi/2]$ , so we need to verify that the sum of our two angles also falls in that range — otherwise the inverse sine on the right might give a different angle (like its supplement). This is a classic pitfall in inverse trigonometric identities.

Let’s denote:

A = \sin^{-1}\frac{8}{17}, \quad B = \sin^{-1}\frac{3}{5}

We want to show $A+B = \sin^{-1}\frac{77}{85}$ .

Find $\cos A$ and $\cos B$ . Since $\sin A = \frac{8}{17}$ and $A$ is in $[-\pi/2, \pi/2]$ , $\cos A \ge 0$ . Using $\sin^2 A + \cos^2 A = 1$ :

\cos A = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}

Similarly, $\sin B = \frac{3}{5}$ , so:

\cos B = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}

Apply the sine addition formula.

\sin(A+B) = \sin A \cos B + \cos A \sin B = \frac{8}{17} \cdot \frac{4}{5} + \frac{15}{17} \cdot \frac{3}{5}

= \frac{32}{85} + \frac{45}{85} = \frac{77}{85}

So $\sin(A+B) = \frac{77}{85}$ . This tells us that $A+B$ is some angle whose sine is $\frac{77}{85}$ . But it could be either $\sin^{-1}\frac{77}{85}$ or $\pi - \sin^{-1}\frac{77}{85}$ , since sine is positive in both first and second quadrants.

Check the range of $A+B$ . We need to see whether $A+B$ $A + B$ lies in $[-\pi/2, \pi/2]$ $[- π /2, π /2]$ (the principal range of $\sin^{-1}$ $sin^{- 1}$ ).
- $A = \sin^{-1}\frac{8}{17}$ . Since $\frac{8}{17} \approx 0.4706$ , $A$ is about $0.49$ radians ( $28^\circ$ ).
- $B = \sin^{-1}\frac{3}{5} \approx 0.6435$ radians ( $36.87^\circ$ ). …

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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 $\tan\dfrac{\theta}{2}=\dfrac{\sin\theta}{1+\cos\theta}=\dfrac{4/5}{1+3/5}=\dfrac{4/5}{8/5}=\dfrac12$ . …
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Q.Prove that $\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}$ .

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

Method: Proving a sum of two inverse sines is another inverse sine

Steps

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Q.Prove that sin⁡−1817+sin⁡−135=sin⁡−17785\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}sin−1178​+sin−153​=sin−18577​.

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Q.Prove that $\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{77}{85}$ .

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