if tan thetA=a/b then the value of a sin theta +b cos theta/a sin theta-b cos theta=
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OpenStudy (anonymous):
theta=x
sinx/cosx=a/b
OpenStudy (anonymous):
a *a+b *b/a*a-b*b
OpenStudy (anonymous):
a^2+b^2/a^2-b^2
OpenStudy (anonymous):
@Mandre
OpenStudy (anonymous):
@theEric
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OpenStudy (souvik):
\[\frac{ a \sin \theta + b \cos \theta }{ a \sin \theta - b \cos \theta }\]
divide both numerator and denominator by \(\cos \theta\) and put the value of \(\tan\theta\)
OpenStudy (anonymous):
a^2/b+1/a^2/b-1
OpenStudy (anonymous):
a^2+b/a^2-b
OpenStudy (anonymous):
same i done above
OpenStudy (souvik):
\[\frac{ \frac{ a^2 }{ b }+b }{ \frac{ a^2 }{ b }-b }\]
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OpenStudy (souvik):
and then..(a^2+b^2)/(a^2-b^2)
OpenStudy (anonymous):
@souvik i have done that earlier see at the top
Parth (parthkohli):
Just because I'm a fan of fancy tricks.\[\dfrac{\sin\theta}{\cos\theta} = \dfrac{a}{b}\]\[\dfrac{a\sin\theta}{b\cos\theta} = \dfrac{a^2}{b^2}\]Applying C&D to both sides,\[\dfrac{a\sin\theta + b\cos\theta}{a\sin\theta - b\cos\theta } = \dfrac{a^2 + b^2}{a^2 - b^2}\]
ganeshie8 (ganeshie8):
clever+cute xD
OpenStudy (anonymous):
after that i have done that earlier in 3rd step
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OpenStudy (anonymous):
see it required two steps.after that
OpenStudy (anonymous):
oh thats the final answer ahh
ganeshie8 (ganeshie8):
yes if u believe in C&D :)
OpenStudy (anonymous):
oh got it before itself.see i used theta =x for my convenience
OpenStudy (anonymous):
theta=x sinx/cosx=a/b
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OpenStudy (anonymous):
a *a+b *b/a*a-b*b
OpenStudy (anonymous):
a^2+b^2/a^2-b^2
OpenStudy (theeric):
You guys all got it, congrats! :) Lots of individual team effort or something :)