Question

please answer this question please​

Answer 1

$\frac{cosecA+1}{cosecA-1}= \frac{1+sinA}{1-sinA}\\ LHS:-\\=\frac{cosecA+1}{cosecA-1}\\=\frac{ \frac{1}{sinA}+1}{ \frac{1}{sinA}-1}\\ =\frac{ \frac{1+sinA}{sinA}}{ \frac{1-sinA}{sinA}}\\= \frac{1+sinA}{sinA}* \frac{sinA}{1-sinA}\\ =\frac{1+sinA}{1-sinA}\\(proved)$

Answer 2

Answer:

$\mathsf{\dfrac{cosecA+1}{cosecA-1}=\dfrac{1+sinA}{1-sinA}}\\\\\\\implies \mathsf{\dfrac{\dfrac{1}{sinA}+1}{\dfrac{1}{sinA}-1}}\\\\\\\implies \mathsf{\dfrac{\dfrac{1+sinA}{sinA}}{\dfrac{1-sinA}{sinA}}}\\\\\\\implies \mathsf{\dfrac{1+sinA}{1-sinA}}$

Step-by-step explanation:

Use the trigonometric identity of :

cosec A = 1 / sin A .

We know that cosec A = (longest side)/(side opposite to A) .

We know that sin A = (side opposite to A)/(longest side).

Hence sin A = 1 / cosec A .

So firstly substitute the value of cosec A to 1 / sin A .

Then simplify further to cancel the denominators and numerators which are same .

We get LHS = RHS and hence the given equation is proved .