Question

Prove that -
sin A
Sec A+tan A-1
cosA
cosecA+cotA-1=1​

Answer 1

Proved below.

Step-by-step explanation:

Given:

$\frac{sin A}{Sec A+tan A-1}+\frac{cosA}{cosecA+cotA-1} =1$

L.H.S.

= $\frac{sin A}{Sec A+tan A-1}+\frac{cosA}{cosecA+cotA-1}$

= $\frac{sinA}{\frac{1}{cosA}+\frac{sinA}{cosA}-1}+\frac{cosA}{\frac{1}{sinA}+\frac{cosA}{sinA}-1 }$       $[secA=\frac{1}{cosA}, tanA= \frac{sinA}{cosA},cosecA=\frac{1}{sinA},cotA=\frac{cosA}{sinA}]$

= $\frac{sinA}{\frac{1+sinA}{cosA}-1 } +\frac{cosA}{\frac{1+cosA}{sinA}-1 }$

= $\frac{sinA}{\frac{1+sinA-cosA}{cosA} } +\frac{cosA}{\frac{1+cosA-sinA}{sinA} }$

= $\frac{sinAcosA}{1+sinA-cosA} +\frac{cosAsinA}{1+cosA-sinA}$

= $sinAcosA[\frac{1}{1+sinA-cosA} +\frac{1}{1+cosA-sinA}]$

= $sinAcosA[\frac{(1+cosA-sinA)+(1+sinA-cosA)}{(1+cosA-sinA)(1+sinA-cosA)}]$

= $sinAcosA[\frac{2}{[1+cosA-sinA][1-(sina-cosA)]} ]$

= $\frac{2sinAcosA}{[1+cosA-sinA][1-(sina-cosA)]}$

= $\frac{2sinAcosA}{1^2-(sinA-cosA)^2}$

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

= $\frac{2sinAcosA}{1-(1-2sinAcosA)}$

= $\frac{2sinAcosA}{1-1+2sinAcosA}$

= $\frac{2sinAcosA}{2sinAcosA}$

= 1

= R.H.S.

Hence proved.