Question

prove that: 1+sinA-cosA=√1-cosA/1+cosA​

Answer 1

Step-by-step explanation:

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

1+sinθ+cosθ

1+sinθ−cosθ

=

1+cosθ

1−cosθ

Step-by-step explanation:

\begin{lgathered}\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\end{lgathered}

1+sinθ+cosθ

1+sinθ−cosθ

=

1+cosθ

1−cosθ

squaring both sides

(

1+sinθ+cosθ

1+sinθ−cosθ

)

2

=

1+cosθ

1−cosθ

solving LHS by using identity

(a+b+c)

2

=a

2

+b

2

+c

2

+2ab+2bc+2ca

(

1+sinθ+cosθ

1+sinθ−cosθ

)

2

\begin{lgathered}\\\\\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)}\end{lgathered}

1+1+2sinθ+2sinθcosθ+2cosθ

1+1+2sinθ−2sinθcosθ−2cosθ

2(1+sinθ)+2cosθ(1+sinθ)

2(1+sinθ)−2cosθ(1+sinθ)

(1+sinθ)(2+2cosθ)

(1+sinθ)(2−2cosθ)

2(1+cosθ)

2(1−cosθ)

(1+cosθ)

(1−cosθ)

Answer 2

Answer:

Step-by-step explanation:

hope it help mark me