Question

1+CotA/Sec A=Sin2A/1-Cos A

Answer 1

Step-by-step explanation:

Given that,

\frac{1 + sec a}{sec a}

seca

1+seca

= \frac{sin 2a}{1 - cos a}

1−cosa

sin2a

Let, L.H.S

\frac{1 + sec a}{sec a}

seca

1+seca

\frac{1 + (\frac{1}{cos a})}{\frac{1}{cos a}}]

cosa

1

1+(

cosa

1

)

]

\frac{\frac{cos a + 1}{cos a}}{\frac{1}{cos a}}

cosa

1

cosa

cosa+1

cos a + 1cosa+1

Multiple and divide with (1 - cos a)

(1 + cos a)\times \frac{1 - cos a}{1 - cos a}(1+cosa)×

1−cosa

1−cosa

\frac{1 - cos^2 a}{1 - cos a}

1−cosa

1−cos

2

a

\frac{sin 2a}{1 - cos a}

1−cosa

sin2a

R.H.S