Question

Please can u solve 1st one

Answer 1

(i)

Here I am writing theta as A.

$Given : \frac{1+cosa+sina}{1+cosa-sina}$

We know that 1 - sina = 1-sin^2a/1+sina

$= \ \textgreater \ \frac{1+cosa+sina}{cosa+ \frac{1-sin^2a}{1+sina} }$

$= \ \textgreater \ \frac{1+cosa+sina}{cosa+ \frac{cos^2a}{1+sina} }$

$= \ \textgreater \ \frac{1+cosa+sina}{ \frac{cosa(sina+1)+cos^2a}{sina+1} }$

$= \ \textgreater \ \frac{(1+cosa+sina)(sina+1)}{cosa(sina+1)+cos^2a}$

$= \ \textgreater \ \frac{(cosa+sina+1)(sina+1)}{cosa(sina+1)+ cos^2a}$

$= \ \textgreater \ \frac{(cosa+sina+1)(sina+1)}{cosa(sina+1)+ cosa*cosa}$

Here, In the denominator I have taken cosa common.

$= \ \textgreater \ \frac{(cosa+sina+1)(sina+1)}{cosa(sina+1+cosa)}$

$= \ \textgreater \ \frac{sina+1}{cosa}$


Hope this helps!