JEE Maths›Inverse Trigonometric Functions›Visual Solutions

Visual Solutions

3 step-by-step animated solutions for Inverse Trigonometric Functions. Each solution breaks the problem into visual clips with key insights.

Tricky4 steps
The value of ∑r=110tan⁡−1(11+r+r2)\displaystyle\sum_{r=1}^{10} \tan^{-1}\left(\dfrac{1}{1+r+r^2}\right) is:
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Tricky4 steps
If α=sin⁡−1(513)\alpha = \sin^{-1}\left(\dfrac{5}{13}\right) and β=cos⁡−1(35)\beta = \cos^{-1}\left(\dfrac{3}{5}\right), then sin⁡(α+β)\sin(\alpha + \beta) equals:
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Hard4 steps
If sin⁡−1(2a1+a2)+cos⁡−1(1−a21+a2)=tan⁡−1(2x1−x2)\sin^{-1}\left(\dfrac{2a}{1+a^2}\right) + \cos^{-1}\left(\dfrac{1-a^2}{1+a^2}\right) = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right), where a,x∈(0,1)a, x \in (0, 1), then the value of xx is:
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