Question

Sec theta - tan theta/ sec theta + tan theta = 1-2 sec theta tan theta +2tan^2 theta

Answer 1

i hope it will help u....

Answer 2

Solution:

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

multiply numerator and denominator by $(sec\theta-tan\theta)$, we get

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

= $\frac{(sec\theta-tan\theta)^{2}}{(sec^{2}\theta)-(tan^{2}\theta)}$

= $\frac{(sec\theta-tan\theta)^{2}}{1}$

/* we know the Trigonometric identity :

sec²A - tan²A = 1 */

= $sec^{2}\theta+tan^{2}\theta-2\times sec\theta \times tan\theta$

=$(1+tan^{2}\theta)+tan^{2}\theta-2\times sec\theta \times tan\theta$

= $1+2tan^{2}\theta-2\times sec\theta \times tan\theta$

= $RHS$

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