document.write( "Question 940703: If $\"+tan%5E2+%2845%2Btheta%29=+a%2Fb+\"$ prove that $\"+%28b-a%29%2F%28b%2Ba%29+=+-sin2theta+\"$ \n" ); document.write( "

Algebra.Com's Answer #573326 by krutarthas(22) $\"\"$ $\"About$

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$\"+tan%5E2+%2845%2Btheta%29=+a%2Fb+\"$
\n" ); document.write( "a/b =tan² (45+θ)
\n" ); document.write( "={tan (45+θ)}²
\n" ); document.write( "={(tan 45+tan θ)/(1-tan 45.tan θ)}²
\n" ); document.write( "={(1+tan θ)/(1-1.tan θ)}²
\n" ); document.write( "=(1+tan θ)²/(1-tan θ)²
\n" ); document.write( "=(1+tan²θ+2tanθ)/(1+tan²θ-2tanθ)
\n" ); document.write( "(b-a)/(b+a)=((1+tan²θ-2tanθ)-(1+tan²θ+2tanθ))/((1+tan²θ-2tanθ)+(1+tan²θ+2tanθ))
\n" ); document.write( "=(1+tan²θ-2tanθ-1-tan²θ-2tanθ)/(1+tan²θ-2tanθ+1+tan²θ+2tanθ)
\n" ); document.write( "=4tanθ/2(1+tan²θ)
\n" ); document.write( "=2tanθ/sec²θ
\n" ); document.write( "=2(sinθ/cosθ)/(1/cos²θ)
\n" ); document.write( "=2sinθ.cos²θ/cosθ
\n" ); document.write( "=2sinθ.cosθ
\n" ); document.write( "=sin2θ \n" ); document.write( "

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