Question

If cosec A=2 then find the value of 1/tanA +sin A/ 1+cos A

Answer 1

We have, cosec A = 2

Now, sin A = $\frac{1}{cosec A }$
⇒ sin A = $\frac{1}{2}$

We know that sin² A + cos² A = 1
⇒ cos² A = 1 - sin² A
⇒ cos A = √(1-sin² A)
⇒ cos A = √{1- $(\frac{1}{2})^{2}$}
⇒ cos A = √(1 - $\frac{1}{4}$)
⇒ cos A = √($\frac{4-1}{4}$)
⇒ cos A = $\frac{ \sqrt{3} }{2}$

We know that tan A = $\frac{sin A}{cos A}$
⇒ tan A = $\frac{1/2}{ \sqrt{3}/2 }$
⇒ tan A = $\frac{1}{ \sqrt{3}}$

We need to find $\frac{1}{tan A}$ + $\frac{sin A}{1 + cos A}$

Substituting the values of each trigonometric function obtained earlier, we get:
$\frac{1}{tan A}$ + $\frac{sin A}{1 + cos A}$ = $\frac{1}{1/ \sqrt{3} }$ + $\frac{1/2}{1 + \sqrt{3}/2 }$
= $\sqrt{3}$ + $\frac{1/2}{(2 + \sqrt{3})/2 }$
= $\sqrt{3}$ + $\frac{1}{2+ \sqrt{3} }$
= $\frac{2 \sqrt{3} + 3 + 1 }{2 + \sqrt{3} }$
= $\frac{2 \sqrt{3} + 4 }{2 + \sqrt{3} }$
= $\frac{2 \sqrt{3} + 3 + 1 }{2 + \sqrt{3} }$ × $\frac{2- \sqrt{3} } {2- \sqrt{3} }$
= $\frac{4 \sqrt{3} - 6 + 8 - 4 \sqrt{3} }{4-3}$ 
= 8-6
= 2

Answer 2

If cosec A=2 then find the value of 1/tanA +sin A/ 1+cos A