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Grade: 11 
        
SecA – CosA =1,then tan2A/2=?
5 years ago

Answers : (1)

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Jitender Singh
IIT Delhi
askIITians Faculty
158 Points 
							
Ans:
secA-cosA = 1
\frac{1}{cosA}-cosA = 1
1-cos^{2}A = cosA
cos^{2}A+ cosA-1=0
cosA = \frac{-1\pm \sqrt{5}}{2}
cosA = \frac{-1+\sqrt{5}}{2}
cosA=1-2sin^{2}\frac{A}{2} = \frac{-1+\sqrt{5}}{2}
2-4sin^{2}\frac{A}{2} = \frac{-1+\sqrt{5}}{1}
4sin^{2}\frac{A}{2}=3-\sqrt{5}
sin^{2}\frac{A}{2}=\frac{3-\sqrt{5}}{4}
cos^{2}\frac{A}{2}=1-sin^{2}\frac{A}{2}=1-\frac{3-\sqrt{5}}{4}=\frac{1+\sqrt{5}}{4}
tan^{2}\frac{A}{2}=\frac{sin^{2}\frac{A}{2}}{cos^{2}\frac{A}{2}}=\frac{3-\sqrt{5}}{1+\sqrt{5}}
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
5 years ago
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