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NCERT Exemplar · Q14

Q.If tan⁡θ=sin⁡α−cos⁡αsin⁡α+cos⁡α\tan\theta = \dfrac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}, then show that sin⁡α+cos⁡α=2 cos⁡θ\sin\alpha + \cos\alpha = \sqrt{2}\,\cos\theta.

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Dividing the given ratio's numerator and denominator by cos⁡α\cos\alpha turns it into tan⁡θ=tan⁡α−1tan⁡α+1=tan⁡(α−π4)\tan\theta=\dfrac{\tan\alpha-1}{\tan\alpha+1}=\tan\left(\alpha-\dfrac{\pi}{4}\right), so θ=α−π4\theta=\alpha-\dfrac{\pi}{4}; substituting α=θ+π4\alpha=\theta+\dfrac{\pi}{4} into sin⁡α+cos⁡α\sin\alpha+\cos\alpha and expanding gives 2cos⁡θ\sqrt{2}\cos\theta.

We are given

tan⁡θ=sin⁡α−cos⁡αsin⁡α+cos⁡α\tan\theta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}

Step 1 — Rewrite the right side in terms of tan⁡α\tan\alpha.

Divide the numerator and denominator by cos⁡α\cos\alpha (valid since sin⁡α+cos⁡α=0\sin\alpha+\cos\alpha=0 would leave tan⁡θ\tan\theta undefined, a case excluded from the statement):

tan⁡θ=sin⁡αcos⁡α−1sin⁡αcos⁡α+1=tan⁡α−1tan⁡α+1\tan\theta=\frac{\frac{\sin\alpha}{\cos\alpha}-1}{\frac{\sin\alpha}{\cos\alpha}+1}=\frac{\tan\alpha-1}{\tan\alpha+1}

Step 2 — Recognise the tangent-subtraction form.

Recall tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}. With A=αA=\alpha, B=π4B=\dfrac{\pi}{4} (so tan⁡B=1\tan B=1):

tan⁡(α−π4)=tan⁡α−11+tan⁡α=tan⁡α−1tan⁡α+1\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{\tan\alpha-1}{1+\tan\alpha}=\frac{\tan\alpha-1}{\tan\alpha+1}

This is exactly our expression, so

tan⁡θ=tan⁡(α−π4)  ⟹  θ=α−π4\tan\theta=\tan\left(\alpha-\frac{\pi}{4}\right) \implies \theta=\alpha-\frac{\pi}{4}

(taking the principal value consistent with the given relation).

Step 3 — Solve for α\alpha and substitute.

α=θ+π4\alpha=\theta+\frac{\pi}{4}

sin⁡α+cos⁡α=sin⁡(θ+π4)+cos⁡(θ+π4)\sin\alpha+\cos\alpha=\sin\left(\theta+\frac{\pi}{4}\right)+\cos\left(\theta+\frac{\pi}{4}\right) …

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