Prove that: sin⁻¹(3/5) + cos⁻¹(5/√26) = tan⁻¹(19/17).

Question

Accepted Answer

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> [!TLDR]
> Convert both inverse-trig terms to a common angle sum and apply the tangent addition formula.
>
> Let $A = \sin^{-1}(3/5)$ and $B = \cos^{-1}(5/\sqrt{26})$. We must show $A+B = 	an^{-1}(19/17)$.
>
> **Angle $A$:** $\sin A = 3/5$. Since $A$ is in $[-\pi/2,\pi/2]$, $\cos A = 4/5$ (positive), so:
> $$	an A = \frac{3}{4}$$
>
> **Angle $B$:** $\cos B = 5/\sqrt{26}$. Since $5^2+1^2=26$, the "opposite side" is $1$, so $\sin B = 1/\sqrt{26}$ (as $B\in[0,\pi]$ and $\cos B>0$ means $B$ is acute), giving:
> $$	an B = \frac15$$
>
> **Adding the angles:**
> $$	an(A+B) = \frac{	an A+	an B}{1-	an A	an B} = \frac{\dfrac34+\dfrac15}{1-\dfrac34\cdot\dfrac15} = \frac{\dfrac{15+4}{20}}{\dfrac{20-3}{20}} = \frac{19/20}{17/20} = \frac{19}{17}$$
> …

Q.Prove that: sin⁻¹(3/5) + cos⁻¹(5/√26) = tan⁻¹(19/17).

Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The core addition identity

Subtraction

The doubling identity

The complementary identity

Q.Prove that: sin⁻¹(3/5) + cos⁻¹(5/√26) = tan⁻¹(19/17).

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