Question

please tell this answer friends please answer 4 th One question only​

Answer 1

Answer:

Step-by-step explanation:

I assume the task is

$\frac{tanA-sinA}{tanA+sinA}$ = $\frac{secA-1}{secA+1}$

L.H. = $\frac{tanA-sinA}{tanA+sinA}$ = $\frac{\frac{sinA-sinAcosA}{cosA} }{\frac{sinA+sinAcosA}{cosA} }$ = $\frac{sinA-sinAcosA}{sinA+sinAcosA}$ = $\frac{sinA(1-cosA)}{sinA(1+cosA)}$ = $\frac{1-cosA}{1+cosA}$

R.H. = $\frac{secA-1}{secA+1}$ = $\frac{\frac{1}{cosA} -1}{\frac{1}{cosA} +1}$ = $\frac{1-cosA}{1+cosA}$

L.H. = R.H.