Question

prove that 1+sin-cos/1+sin+cos=√1-cos/√1+cos​

Answer 1

ANSWER:

To Prove:

   $\:\:\:\:\bullet\:\:\:\:\sf{\dfrac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}=\dfrac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}}$

Proof:

$:\longrightarrow\sf{\dfrac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}=\dfrac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}}\\\\\text{\sf{Solving LHS,}}\\\\:\implies\sf{\dfrac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}}\\\\\text{\sf{On Squaring LHS,}}\\\\:\implies\sf{\left(\dfrac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\right)^2}$

$\text{\sf{We know that, (a+b+c) ^2 = a ^2 +b ^2 +c ^2 +2ab+2bc+2ca.}}\\\\\text{\sf{So,}}\\\\:\implies\sf{\left[\dfrac{1+\sin^2\theta+\cos^2\theta+2\sin\theta-2\cos\theta-2\sin\theta\cos\theta}{1+\sin^2\theta+\cos^2\theta+2\sin\theta+2\cos\theta+2\sin\theta\cos\theta}\right]}\\\\\text{\sf{We know that, \sin^2\theta+\cos^2\theta=1 }}\\\\\text{\sf{So,}}\\\\:\implies\sf{\left[\dfrac{1+1+2\sin\theta-2\cos\theta-2\sin\theta\cos\theta}{1+1+2\sin\theta+2\cos\theta+2\sin\theta\cos\theta}\right]}$

$:\implies\sf{\left[\dfrac{2+2\sin\theta-2\cos\theta-2\sin\theta\cos\theta}{2+2\sin\theta+2\cos\theta+2\sin\theta\cos\theta}\right]}\\\\\text{\sf{Taking 2 common in both numerator and denominator,}}\\\\:\implies\sf{\dfrac{2\!\!\!/\times(1+\sin\theta-\cos\theta-\sin\theta\cos\theta)}{2\!\!\!/\times(1+\sin\theta+\cos\theta+\sin\theta\cos\theta)}}\\\\:\implies\sf{\dfrac{1+\sin\theta-\cos\theta-\sin\theta\cos\theta}{1+\sin\theta+\cos\theta+\sin\theta\cos\theta}}\\\\\text{\sf{Taking \cos\theta common,}}$

$:\implies\sf{\dfrac{(1+\sin\theta)-\cos\theta(1+\sin\theta)}{(1+\sin\theta)+\cos\theta(1+\sin\theta)}}\\\\\text{\sf{Taking 1+\sin\theta common,}}\\\\:\implies\sf{\dfrac{(1+\sin\theta)\times(1-\cos\theta)}{(1+\sin\theta)\times(1+\cos\theta)}}\\\\:\implies\sf{\dfrac{1-\cos\theta}{1+\cos\theta}}\\\\\text{\sf{Because we had squared the LHS earlier, now we take its square root,}}\\\\\text{\sf{Hence,}}$

$\bf{:\implies\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}=\dfrac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}=RHS}\\\\\text{HENCE PROVED}$

Formulae Used:

• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• sin²Ф + cos²Ф = 1