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Grade: 11

                        

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

one year ago

Answers : (2)

Arun
24737 Points
							
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)
one year ago
Aditya Gupta
2056 Points
							
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 :)
one year ago
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