Start off by using

tan^2(A/2) = ( sin^2(A)/(1+cosA)^2)

So tan^2(x/2) = ( sin^2(x)/(1+cosx)^2)

= ( 1- cos^2(x))/(1+cosx)^2

= (1-cosx)(1+cosx)/((1+cosx)^2)

Cancelling term gives

(1-cosx)/(1+cosx)

= (secx-1)/(1+secx)

LHS=RHS

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Dr. Badi M
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2 years ago

When proving such relations we have a choice, that is we can either start from the left hand side (LHS) and show that it equals the right hand side (RHS) or vice versa.

I will start with the RHS,

RHS = tan^2(x/2),

= {sin^2(x/2)} / {cos^2(x/2)},

= {(1/2)(1+cos(x))} / {(1/2)(1-cos(x))}

= {1+cos(x)} / {1-cos(x)},

= {1+(1/sec(x))} / {1-(1/sec(x))},... more

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Raafay Karim
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38 students helped

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3 years ago

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