1nice nd simple soln....
Reduce to simplest form tan^{-1}(\frac {xcos\theta}{1-xsin\theta})-cot^{-1}(\frac {cos\theta}{x-sin\theta})
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5 Answers
1tan-1(xcosθ/1-xsinθ)
Take x=sin θ
=tan-1(sinθcosθ/1-sin2θ)
=tan-1(sinθcosθ/cos2θ)
=tan-1(tanθ)
=θ=sin-1x
tan-1(xcosθ/1-xsinθ)=sin-1x (1)
cot-1(cosθ/x-sinθ)
Take x=cosecθ
=cot-1(cosθ/(1/sinθ)-sinθ
=cot-1(cosθsinθ/1-sin2θ)
=cot-1(cosθsinθ/cos2θ)
=cot-1(tanθ)
=cot-1[cot((Π/2) -θ)]
=(Π/2)-θ =(Π/2)-cosec-1x
cot-1(cosθ/x-sinθ)=sec-1x (2)
putting (1) and (2) in the given equation,we get
tan-1(xcosθ/1-xsinθ) - cot-1(cosθ/x-sinθ)=sin-1x - sec-1x
=sin-1x- cos-1(1/x)
wrong soln...
62Honey, can you take x=sin theta?
no .. because the variable is already used
if at all you can use x= sin alpha! (Taht too after due diligence)
4Given expression = tan-1 (x / (secθ - x tanθ) ) - tan-1(x secθ - tanθ)
= tan-1 (x/ secθ -x tanθ) - (x secθ - tanθ)] / [ 1 + (x secθ - tanθ)(x/(secθ - x tanθ) ]
= tan-1 [tanθ ]
= θ