Question

if cot theta =15/8 then evaluate (1+sin theta) (1-sin theta) / ( 1+cos theta) (1-cos theta)​

Answer 1

$\huge\bold{\underbrace{\red{SOLUTION:-}}}$

GIVEN :-

• $\rm{\cot\theta\:=\:\dfrac{15}{8}\:}$

FORMULA :-

• $\rm{\green{{(a\:+\:b)}{(a\:-\:b)}\:=\:a^2\:-\:b^2\:}}$----(1)

• $\rm{\red{\dfrac{\sin\theta}{\cos\theta}\:=\:\tan\theta\:}}$

• $\rm{\orange{\dfrac{\cos\theta}{\sin\theta}\:=\:\cot\theta\:}}$

• $\rm{\blue{\cosec^2\theta\:-\:\cot^2\theta\:=\:1\:}}$

• $\rm{\implies\:\cosec^2\theta\:-\:1\:=\:\cot^2\theta\:}$

CALCULATION :-

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

✍️ Here, in numerator

• a = 1 and b = sinθ

✍️ In denominator,

• a = 1 and b = cosθ

✍️ Now putting the formula (1), we get

$= \frac{1 - { \sin^{2} \theta } }{1 - { \cos^{2} \theta } }$

✍️ Now, dividing ‘sin²θ’ in both numerator and denominator .

$= \frac{ \frac{1 - { \sin^{2} \theta } }{ { \sin^{2} \theta } } }{ \frac{1 - { \cos^{2} \theta } }{ { \sin^{2} \theta } } } \\ \\ = \frac{ \frac{1}{ { \sin^{2} \theta } } - \frac{ { \sin^{2} \theta } }{ { \sin^{2} \theta } } }{ \frac{1}{ { \sin^{2} \theta } } - \frac{ { \cos^{2} \theta } }{ { \sin^{2} \theta } } } \\ \\ = \frac{ { \cosec^{2} \theta } - 1 }{ { \cosec^{2} \theta } - { \cot^{2} \theta } } \\ \\ = \frac{ { \cot^{2} \theta } }{1} \\ \\ = {( \cot \theta )}^{2}$

✍️ Now put the value of ‘cotθ’ , we get

$\rm{=\:(\dfrac{15}{8})^2\:}$

$\rm{=\:\dfrac{225}{64}\:}$

$\bigstar\:\mathcal{\purple{\boxed{\dfrac{(1\:+\:\sin\theta)\:(1\:-\:\sin\theta)}{(1\:+\:\cos\theta)\:(1\:-\:\cos\theta)}\:=\:\dfrac{225}{64}\:}}}$