Question

sec theta-tan theta upon sec theta + tan theta= 1-2sec theta tan theta +2 tan square theta

Answer 1

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Answer 2

Answer:

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

Step-by-step explanation:

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

/* By Trigonometric identity:

sec²A-tan²A = 1 */

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

Therefore,

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

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