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Kindly tell last question...value of tan alpha..........plxxx see image

Kindly tell last question...value of tan alpha..........plxxx see image

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Grade:11

2 Answers

Arun
25750 Points
4 years ago
Dear student
 
I have solved reverse.
 
LHS = \frac{1-2tan^{2}\beta}{1+2tan^{2}\beta}LHS = \frac{2-4tan^{2}\beta}{2+4tan^{2}\beta}LHS = \frac{3-3tan^{2}\beta-1-tan^{2}\beta }{3+3tan^{2}\beta-1+tan^{2}\beta }LHS = \frac{3.\frac{1-tan^{2}\beta }{1+tan^{2}\beta } -1}{3-\frac{1-tan^{2}\beta }{1+tan^{2}\beta } }LHS = \frac{3cos2\beta -1}{3-cos2\beta }RHS = cos2\alpha = \frac{1-2tan^{2}\beta }{1+2tan^{2}\beta }RHS = cos2\alpha = \frac{1-(\sqrt{2}tan\beta )^{2} }{1+(\sqrt{2}tan\beta )^{2}}cos2\alpha = \frac{1-tan^{2}\alpha }{1+tan^{2}\alpha }tan\alpha = \sqrt{2}tan\betaquestion
 
Hope it helps
In case of any difficulty, please feel free to ask again.
 
Regards
Arun (askIITians forum expert)
Aditya Gupta
2081 Points
4 years ago
hello tanisha this ques can be easily solved by applying two things: componendo and dividendo, and the formula cos2x= (1 – t^2)/(1+t^2) where t= tanx
so the given expression is 
(1 – t^2)/(1+t^2)= (3cos2b – 1)/(3 – cos2b) where t= tanalpha
now apply C n D.
 – 2t^2/2= 4(cos2b – 1)/2(cos2b+1)
or t^2= 2(1 – cos2b)/(cos2b+1)= 2tan^2b (here we have also used (1 – cos2x)/(1+cos2x)= tan^2x)
hence t= tanalpha= ± tanb*root2
so both options (1) and (2) are correct.
kndly approve :)

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