Question

NCERT class 10 Introduction to trigonometry example 15

Answer 1

Answer:

Step-by-step explanation:

In pic

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Hope it helped.

Answer 2

Answer:

To Prove: $\frac{sin\,\theta-cos\,\theta+1}{sin\,\theta+cos\,\theta-1}=\frac{1}{sec\,\theta-tan\,\theta}$

Consider,

LHS

$=\frac{sin\,\theta-cos\,\theta+1}{sin\,\theta+cos\,\theta-1}$

Divide numerator and denominator by $cos\,\theta$

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

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

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

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

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

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

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

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

= RHS

Hence Proved