Question

prove that sin theta/1-cos theta + tan theta/1+cos theta=sec theta *cosec theta + cot theta

Answer 1

step by step explanation

Answer 2

Answer:

We have to prove the given Expression:

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

Consider,

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

Making Denominator same,

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

using multiplication in numerator and identity in denominator

( a² - b² = ( a - b ) ( a + b ) )

$=\frac{sin\,\theta+sin\,\theta\:cos\,\theta+tan\,\theta-tan\,\theta\:cos\,\theta}{1-cos^2\,\theta}$

$=\frac{sin\,\theta+sin\,\theta\:cos\,\theta+tan\,\theta-\frac{sin\,\theta}{cos\,\theta}\times cos\,\theta}{1-cos^2\,\theta}$

using identity $sin^2\,\theta+cos^2\,\theta=1$

$=\frac{sin\,\theta+sin\,\theta\:cos\,\theta+tan\,\theta-sin\,\theta}{sin^2\,\theta}$

$=\frac{sin\,\theta\:cos\,\theta+tan\,\theta}{sin^2\,\theta}$

$=\frac{sin\,\theta\:cos\,\theta}{sin^2\,\theta}+\frac{tan\,\theta}{sin^2\,\theta}$

$=\frac{cos\,\theta}{sin\,\theta}+\frac{\frac{sin\,\theta}{cos\,\theta}}{sin^2\,\theta}$

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

$=cot\,\theta+cosec\,\theta\:sec\,\theta$

$=RHS$

Hence Proved