PROVE THE FOLLOWING IDENTITIES.

LHS=2t^2/1-t^2

RHS=2t^2/1-t^2

Hence left had side is equal

Step-by-step explanation:

Taking tan x and sec x in the terms of t=tan(x/2)

tan x=2t/1-t^2

sex x=1/cos x

therefore cos x=1-t^2/1+t^2

so sec x=1/1-t^2/1+t^2

sec x=1+t^2/1-t^2

Taking tan x and sec x in the terms of t=tan(x/2)

tan x=2t/1-t^2

sex x=1/cos x

therefore cos x=1-t^2/1+t^2

so sec x=1/1-t^2/1+t^2

sec x=1+t^2/1-t^2

LHS=tanx .tan(x/2)

      =2t.t/1-t^2

      =2t^2/1-t^2

RHS=sec x-1

       =1+t^2/1-t^2-1

       =1+t^2-1+t^2/1-t^2

       =2t^2/1-t^2

Therefore hence we proved LHS=RHS

      =2t.t/1-t^2

      =2t^2/1-t^2

RHS=sec x-1

       =1+t^2/1-t^2-1

       =1+t^2-1+t^2/1-t^2

       =2t^2/1-t^2

Therefore hence we proved LHS=RHS

Step-by-step explanation:

tan x tan(x/2)

Use half angle formula.

tan x ((1 − cos x) / sin x)

Distribute.

(tan x − sin x) / sin x

Divide.

sec x − 1

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