Question

1/sec theta - tan theta = sec theta+tan theta

Answer 1

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Answer 2

Answer:

$\frac{1}{sec\theta-tan\theta}=sec\theta + tan\theta$

Step-by-step explanation:

$L.H.S = \frac{1}{sec\theta-tan\theta}\\=\frac{(sec\theta+tan\theta)}{(sec\theta-tan\theta)(sec\theta+tan\theta)}\\=\frac{(sec\theta+tan\theta)}{sec^{2}\theta-tan^{2}\theta}$

/* By algebraic identity:

(a+b)(a-b)=a²-b² */

$=\frac{(sec\theta+tan\theta)}{1}$

/* By Trigonometric identity:

sec²A - tan²A = 1 */

$=sec\theta + tan\theta$

$=R.H.S$

Therefore,

$\frac{1}{sec\theta-tan\theta}=sec\theta + tan\theta$