Question

prove that 1/sec theta+tan theta =sec theta -tan theta.
please answer in full step

Answer 1

Answer:

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

Step-by-step explanation:

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

$By \: trigonometric \: identity :\\</p><p>\boxed { sec^{2}\theta - tan^{2}\theta }$

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

/* By algebraic identity:

a² - b² = ( a + b )( a - b ) */

$= sec\theta-tan\theta\\=RHS$

Therefore,.

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

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Answer 2

Answer:

LHS=ee33hfgfftyyy is a very good