Question

solve the following​

Answer 1

Step-by-step explanation:

let theta = x

so we have,

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

LHS,

(sinx/cosx)/(1/cosx+1)

= sinx/(1+cosx)

= (2sinx/2cosx/2)/2cos²x/2

as, sin2A = 2sinAcosA

and 1+cos2A = 2cos²A

so, (sinx/2)/(cosx/2)

=> tanx/2

LHS,

(1/cosx-1)/(sinx/cosx)

= (1-cosx)/sinx

= (2sin²x/2)/(2sinx/2cosx/2)

as, 1-cos2A = 2sin²A

so, (sinx/2)/(cosx/2)

=> tanx/2

so, LHS = RHS

hence proved....

OR

as we have,

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

cross-multiplication

tan²x = (secx-1)(secx+1)

tan²x = sec²x-1

1 = sec²x-tan²x

as it is identity that,,

sec²x-tan²x = 1

so in above eq.

1=1

i.e., LHS = RHS

hence proved....