Question

cotA =7/8 than (1+ sinA)(1- sinA)/(1+ cosA)(1- cosA)

Answer 1

i hope this will help u

Answer 2

Given:

$\cot A$ = $\dfrac{7}{8}$

We have to find, the value of $\dfrac{(1+ \sin A)(1- \sin A)}{(1+ \cos A)(1- \cos A)}$ is:

Solution:

∴ $\dfrac{(1+ \sin A)(1- \sin A)}{(1+ \cos A)(1- \cos A)}$

Using the algebraic identity:

(a + b)(a - b) = $a^{2} -b^{2}$

$=\dfrac{1^2- \sin^2 A}{1^2- \cos^2 A}$

$=\dfrac{1- \sin^2 A}{1- \cos^2 A}$

Using the trigonometric identity:

$\sin^2 A$ + $\cos^2 A$ = 1

⇒ $\cos^2 A$ = 1 - $\sin^2 A$ and $\sin^2 A$ = 1 - $\cos^2 A$

$=\dfrac{\cos^2 A}{ \sin^2 A}$

$=(\dfrac{\cos A}{ \sin A})^2$

Using the trigonometric identity:

$\cot A$ = $\dfrac{\cos A}{ \sin A}$

= $\cot^2 A$

Put $\cot A$ = $\dfrac{7}{8}$, we get

= $(\dfrac{7}{8})^2$

= $\dfrac{49}{64}$

∴ $\dfrac{(1+ \sin A)(1- \sin A)}{(1+ \cos A)(1- \cos A)}$ = $\dfrac{49}{64}$

Thus, the value of $\dfrac{(1+ \sin A)(1- \sin A)}{(1+ \cos A)(1- \cos A)}$ is "equal to $\dfrac{49}{64}$".