Question

if sectheta-tantheta=k,then prove that(k²+1)sintheta=(k²-1)​

Answer 1

Given : sectheta-tantheta=k

secθ - tanθ = k

To Find : prove that (k²+1)sin θ=(k²-1)​

Solution:

secθ - tanθ = k

As we know that

sec²θ - tan²θ =1

=> (secθ +  tanθ )(secθ - tanθ ) = 1

=> (secθ +  tanθ )(k ) = 1

=> secθ +  tanθ = 1/k

secθ - tanθ = k

secθ +  tanθ = 1/k

=> 2secθ = k + 1/k

=>  secθ = (k² + 1)/2k

=> cosθ = 2k/ (k² + 1)

2tanθ = 1/k - k

=> tanθ  = ( 1- k²)/2k

(k²+1)sin θ=(k²-1)​

LHS = (k²+1)sin θ

=  (k²+1) cosθtan θ

=  (k²+1) (2k/ (k² + 1)) (( 1- k ²_/2k)

=  1- k ²

≠RHS

problem in data

if secθ + tanθ = k  then it will satisfy as then tanθ  = (  k² - 1 )/2k

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