Question

Show that (sin€+1+cos€) (sin€-1+cos€) *Sec€*cosec€=2

Answer 1

Answer:

Step-by-step explanation:

Let us assume theta to be "A"

To prove:

(sinA+1+cosA)(sinA–1+cosA)(secAcosecA) = 2

Proof

LHS

(sin^2A –sinA+sinAcosA+sinA–1+cosA+cosAsinA–cosA+cos^2A)(1/sinAcosA)

sin^2A+cos^2A–1–sinA+sinA+cosA–cosA+sinAcosA+cosAsinA)(1/sinAcosA)

(1–1+2sinAcosA)(1/sinAcosA)

2sinAcosA/cosAsinA

= 2

LHS = RHS

Hence proved

Hope it helps you

Answer 2

Answer

PROVE THAT : (sin€+cos€+1)(sin€+cos€-1)sec€cosec€=2