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NCERT Exemplar · Q68

Q.If tan⁡A=1−cos⁡Bsin⁡B\tan A = \dfrac{1 - \cos B}{\sin B}, then tan⁡2A=tan⁡B\tan 2A = \tan B.

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The key idea is to rewrite the given expression for tan⁡A\tan A using half-angle identities, which directly shows A=B2A = \frac{B}{2} (or a related form), and then tan⁡2A=tan⁡B\tan 2A = \tan B follows immediately.

We are given:

tan⁡A=1−cos⁡Bsin⁡B\tan A = \frac{1 - \cos B}{\sin B}

The goal is to prove that tan⁡2A=tan⁡B\tan 2A = \tan B. The natural instinct is to try to simplify the right-hand side into something like tan⁡(something)\tan(\text{something}), so that we can relate AA to BB.


Why this approach works

The expression 1−cos⁡Bsin⁡B\frac{1 - \cos B}{\sin B} is a classic half-angle form. Recall the identities:

1−cos⁡B=2sin⁡2B2,sin⁡B=2sin⁡B2cos⁡B21 - \cos B = 2 \sin^2 \frac{B}{2}, \quad \sin B = 2 \sin \frac{B}{2} \cos \frac{B}{2}

So the fraction simplifies to tan⁡B2\tan \frac{B}{2}. That means tan⁡A=tan⁡B2\tan A = \tan \frac{B}{2}, which implies A=B2+nπA = \frac{B}{2} + n\pi (for integer nn). Then tan⁡2A=tan⁡B\tan 2A = \tan B follows directly.

Let’s do it step by step.


  1. Rewrite numerator and denominator using half-angle formulas We know:

1−cos⁡B=2sin⁡2B21 - \cos B = 2 \sin^2 \frac{B}{2}

sin⁡B=2sin⁡B2cos⁡B2\sin B = 2 \sin \frac{B}{2} \cos \frac{B}{2}

Substitute into the given:

tan⁡A=2sin⁡2B22sin⁡B2cos⁡B2\tan A = \frac{2 \sin^2 \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}}

  1. Cancel common factors Provided sin⁡B2≠0\sin \frac{B}{2} \neq 0 (which is fine for general BB except multiples of 2π2\pi), we cancel 2sin⁡B22 \sin \frac{B}{2}:

tan⁡A=sin⁡B2cos⁡B2=tan⁡B2\tan A = \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}} = \tan \frac{B}{2}

Tip

This cancellation is valid as long as sin⁡B2≠0\sin \frac{B}{2} \neq 0. If sin⁡B2=0\sin \frac{B}{2} = 0, then B=2nπB = 2n\pi, making tan⁡B=0\tan B = 0 and the original expression for tan⁡A\tan A becomes 0/00/0 (indeterminate), so we exclude that degenerate case.

  1. Interpret the equality of tangents …

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