Question

Prove that:
(1 + cot A - cosec A) (1 + tan A + sec A) = 2

Answer 1

[tex](1+ cot A - cosec A)(1+tanA+secA)=2 \\ L.H.S. \\ =(1+cotA-cosecA)(1+tanA+secA) \\ = (1 + \frac{cosA}{sinA} - \frac{1}{sinA} )(1+ \frac{sinA}{cosA} + \frac{1}{cosA} ) \\ = ( \frac{sinA+cosA-1}{sinA} )( \frac{cosA+sinA+1}{cosA} ) \\ = \frac{ (sinA+cosA)^{2}- 1^{2} }{sinA.cosA} \\ = \frac{ sin^{2}A + cos^{2}A + 2sinA.cosA-1 }{sinA.cosA} \\ = \frac{ 1 + 2sinA.cosA-1 }{sinA.cosA} \\ = \frac{2sinA.cosA }{sinA.cosA} = 2 = R.H.S.[/tex]
Hence, proved.

Answer 2

(1+cot A-cosec A).(1+tanA+secA)= 2

L.H.S.

=(1+cosA/sinA-1/sinA).(1+sinA/cosA+1/cosA)

=(sinA+cosA-1)×(cosA+sinA+1)/sinA.cosA

=[(sinA+cosA)^2-(1)^2]/sinA.cosA.

=(sin^2A+cos^2A+2.sinA.cosA-1)/sinA.cosA.

=( 1+2.sinA.cosA -1)/sinA.cosA.

= 2.sinA.cosA/sinA.cosA

= 2 , proved.