Question

sinA - CosoA +1
SinoA + CosoA -1=1/(secA - tanA)​

Answer 1

Answer:

cos A = 1-sinA this easy not so hard

Answer 2

$\dfrac{\sin A-\cos A +1}{\sin A+\cos A-1} =\dfrac{1}{\sec A-\tan A}$, proved.

Step-by-step explanation:

Prove that, $\dfrac{\sin A-\cos A +1}{\sin A+\cos A-1} =\dfrac{1}{\sec A-\tan A}$.

L.H.S. $=\dfrac{\sin A-\cos A +1}{\sin A+\cos A-1}$

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

Rationalising numerator and denominator, we get

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

$=\dfrac{[\sin A-(\cos A -1)]^2}{\sin^2 A-(\cos A-1)^2}$

$=\dfrac{[\sin^2 A+(\cos A -1)]^2+2\sin A(\cos A -1)}{\sin^2 A-\cos^2 A-1+2\cos A}$

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

= $\dfrac{1+\sin A}{\cos A}$

$=\sec A+\tan A$

R.H.S. $=\dfrac{1}{\sec A-\tan A}$

$=\dfrac{1}{\sec A-\tan A}\times \dfrac{\sec A+\tan A}{\sec A+\tan A}$

$=\dfrac{\sec A+\tan A}{\sec^2 A-\tan^2 A}$

Using the trigonometric identity,

$\sec^2 A-\tan^2 A=1$

$=\sec A+\tan A$

∴ L.H.S. = R.H.S. $=\sec A+\tan A$, proved.