Question

Prove that sinA/secA + tanA - 1 + cosA/cosecA + cotA - 1 = 1

Answer 1

sinA/secA+tanA-1+cosA/cosecA+cotA-1
=sinA/(1/cosA+sinA/cosA-1)+cosA/(1/sinA+cosA/sinA-1)
=sinA/{(1+sinA-cosA)/cosA}+cosA/{(1+cosA-sinA)/sinA}
=sinAcosA/(1+sinA-cosA)+sinAcosA/(1+cosA-sinA)
=sinAcosA[(1+cosA-sinA+1+sinA-cosA)/(1+sinA-cosA)(1+cosA-sinA)]
=2sinAcosA/(1+sinA-cosA+cosA+sinAcosA-cos²A-sinA-sin²A+sinAcosA)
=2sinAcosA/{1+2sinAcosA-(sin²A+cos²A)}
=2sinAcosA/(1+2sinAcosA-1)
=2sinAcosA/2sinAcosA
=1 (Proved)

Answer 2

Given that,

The function is

$\dfrac{\sin A}{\sec A+\tan A-1}+\dfrac{\cos A}{cosec A+cot A-1}=1$

We need to prove the left hand side equal right hand side

Using left hand side

$=\dfrac{\sin A}{\sec A+\tan A-1}+\dfrac{\cos A}{cosec A+cot A-1}$

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

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

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

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

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

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

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

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

Here, $\sin^{2}A+\cos^{2}A=1$

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

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

$=1$

This is right hand side

Hence, it is proved.