Question

1+sinA-cosA/1+sinA+cosA=√1-cosA/1+cosA​

Answer 1

Step-by-step explanation:

⭐LHS = [(1+sinA-cosA )/(1+sinA+cosA)]²

=[ (1+sin²A+cos²A+2sinA-2sinAcosA-2cosA)/(1+sin²A+cos²A+2sinA+2sinAcosA+2cosA) ]²

= [( 1+1+2sinA-2sinAcosA-2cosA)/(1+1+2sinA+2sinAcosA+2cosA)]²

= { ( 2(1+sinA) -2cosA(sinA+1) / [ 2(1+sinA) + 2cosA(sinA +1 ) ]}²

= { (2(1+sinA)[1 - cosA] /[2(1+sinA )(1 + cosA )] }²

= [ ( 1 - cosA ) / ( 1 + cosA ) ]²

⭐= RHS

#Regards♥️

Answer 2

Given : $\frac{1+\sin\theta-\cos \theta}{1+\sin\theta+\cos \theta}= \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$

Step-by-step explanation:

$\frac{1+\sin\theta-\cos \theta}{1+\sin\theta+\cos \theta}= \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\\\\\text{squaring both sides }\\\\(\frac{1+\sin\theta-\cos \theta}{1+\sin\theta+\cos \theta})^2=\frac{1-\cos\theta}{1+\cos\theta}\\\\\text{solving LHS by using identity} \\\\(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\\\\(\frac{1+\sin\theta-\cos \theta}{1+\sin\theta+\cos \theta})^2$

$\\\\\frac{1+1+2\sin\theta-2\sin\theta\cos\theta-2\cos\theta}{1+1+2\sin\theta+2\sin\theta\cos\theta+2\cos\theta}\\\\\frac{2(1+\sin\theta)-2\cos\theta(1+\sin\theta)}{2(1+\sin\theta)+2\cos\theta(1+\sin\theta)}\\\\\frac{(1+\sin\theta)(2-2\cos\theta)}{(1+\sin\theta)(2+2\cos\theta)}\\\\\frac{2(1-\cos\theta)}{2(1+\cos\theta)}\\\\\frac{(1-\cos\theta)}{(1+\cos\theta)}$

hence proved

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