Question

If tan theta + sin theta = m and tan theta - sin theta = n prove that(m square minus n square) square =16mn

Answer 1

igstiotjsgkzgzotsyzyidiyx

Answer 2

m=tanΘ+sinΘ
On squaring,
m^2=(tanΘ+sinΘ)^2
       =tan^2Θ+sin^2Θ+2tanΘ·sinΘ  
n=tanΘ-sinΘ
On squaring,
n^2=(tanΘ-sinΘ)^2
      =tan^2Θ+sin^2Θ-2tanΘ·sinΘ  
To prove (m^2-n^2)^2=16mn
LHS
 (m^2-n^2)^2
On substituting,
[(tan^2Θ+sin^2Θ+2tanΘ·sinΘ)- (tan^2Θ+sin^2Θ-2tanΘ·sinΘ )]^2
[tan^2Θ+sin^2Θ+2tanΘ·sinΘ-tan^2Θ-sin^2Θ+2tanΘ·sinΘ]^2  
[2tanΘ·sinΘ+2tanΘ·sinΘ]^2
[4tanΘ·sinΘ]^2
[16·tan^2Θ·sin^2Θ]
RHS
16mn
16[(tanΘ+sinΘ)(tanΘ-sinΘ)]
16[(tan^2Θ-sin^2Θ)]
16[(sin^2Θ/cos^2Θ)-sin^2Θ]
16{[sin^2Θ(1-Θcos^2Θ)]/cos^2Θ}
16{[sin^2Θ·(sin^2Θ)]/cos^2Θ}          [Because sin^2Θ+cos^2Θ=1]
16[sin^2Θ·tan^2Θ]     
[16·tan^2Θ·sin^2Θ]
Therefore LHS=RHS
So (m^2-n^2)^2=16mn