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Grade: 12th pass 

                        
Dear Sir, Please solve the problem attached here with...no.6 only

							
Dear Sir,
 
Please solve the problem attached here with...no.6 only
3 years ago

Answers : (1)

Deep Chand
42 Points 
							
If sin\theta, cos\theta and tan\theta are in  G.P., then
cos^2\theta=sin\theta.tan\theta
cos^2\theta=sin\theta.\frac {sin\theta}{cos\theta}
cos^3\theta=sin^2\theta
Squaring both side we get
cos^6\theta=sin^4\theta
Multiply and Divide by sin^2 \theta in Numerator and denominator on RHS
cos^6\theta=sin^4\theta.\frac {sin^2\theta}{sin^2\theta}
\frac{cos^6\theta}{sin^6\theta}=cosec^2\theta
cot^6\theta=1+cot^2\theta
cot^6\theta -cot^2\theta=1
Proved
DEEP CHAND SAROJ 
B.Tech (IIT -DELHI)
3 years ago
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