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NCERT Exemplar · Q1

Q.Prove that tan⁡A+sec⁡A−1tan⁡A−sec⁡A+1=1+sin⁡Acos⁡A\dfrac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \dfrac{1 + \sin A}{\cos A}.

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Using sec⁡2A−tan⁡2A=1\sec^2A-\tan^2A=1, write 1=(sec⁡A−tan⁡A)(sec⁡A+tan⁡A)1=(\sec A-\tan A)(\sec A+\tan A); this lets the numerator factor to cancel exactly with the denominator, leaving sec⁡A+tan⁡A=1+sin⁡Acos⁡A\sec A+\tan A=\dfrac{1+\sin A}{\cos A}.

We start from the Pythagorean identity sec⁡2A−tan⁡2A=1\sec^2A-\tan^2A=1, which factors as

1=(sec⁡A−tan⁡A)(sec⁡A+tan⁡A)1=(\sec A-\tan A)(\sec A+\tan A)

Step 1 — Rewrite the numerator.

In the numerator tan⁡A+sec⁡A−1\tan A+\sec A-1, replace the −1-1 with −(sec⁡A−tan⁡A)(sec⁡A+tan⁡A)-(\sec A-\tan A)(\sec A+\tan A):

tan⁡A+sec⁡A−1=(sec⁡A+tan⁡A)−(sec⁡A−tan⁡A)(sec⁡A+tan⁡A)\tan A+\sec A-1=(\sec A+\tan A)-(\sec A-\tan A)(\sec A+\tan A)

Factor out (sec⁡A+tan⁡A)(\sec A+\tan A):

=(sec⁡A+tan⁡A)[1−(sec⁡A−tan⁡A)]=(sec⁡A+tan⁡A)(tan⁡A−sec⁡A+1)=(\sec A+\tan A)\big[1-(\sec A-\tan A)\big]=(\sec A+\tan A)(\tan A-\sec A+1)

Step 2 — Divide by the denominator.

The denominator is exactly tan⁡A−sec⁡A+1\tan A-\sec A+1, so it cancels:

tan⁡A+sec⁡A−1tan⁡A−sec⁡A+1=(sec⁡A+tan⁡A)(tan⁡A−sec⁡A+1)tan⁡A−sec⁡A+1=sec⁡A+tan⁡A\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{(\sec A+\tan A)(\tan A-\sec A+1)}{\tan A-\sec A+1}=\sec A+\tan A

Step 3 — Convert to sine and cosine.

sec⁡A+tan⁡A=1cos⁡A+sin⁡Acos⁡A=1+sin⁡Acos⁡A\sec A+\tan A=\frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}

This is exactly the right-hand side.

✓Final answer

tan⁡A+sec⁡A−1tan⁡A−sec⁡A+1=1+sin⁡Acos⁡A\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1+\sin A}{\cos A} — proved.

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