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Prove that
{1/(sec^2 a-cos^2 a) + 1/(cosec^2 a-sin^ a)}*cos^2 a sin^2 a=(1-cos^2 a sin^2 a)/(2+cos^2 a sin^2 a)

Bahirithi , 7 Years ago
Grade 12
anser 1 Answers
Aarushi Ahlawat
 
 
First convert all the terms to sin and cos terms and then use identity to replace sin^2a+cos^a with 1. Here are the steps
LHS=\left [ \frac{1}{sec^{2}(a)-cos^{2}(a)}+\frac{1}{cosec^{2}(a)-sin^{2}(a)} \right ]cos^{2}(a)sin^{2}(a)
 
=\left [ \frac{cos^{2}(a)}{1-cos^{4}(a)}+\frac{sin^{2}(a)}{1-sin^{4}(a)} \right ]cos^{2}(a)sin^{2}(a)
=\left [ \frac{cos^{2}(a)}{(1-cos^{2}(a))(1+cos^{2}(a))}+\frac{sin^{2}(a)}{(1-sin^{2}(a))(1+sin^{2}(a))} \right ]cos^{2}(a)sin^{2}(a)
=\left [ \frac{cos^{2}(a)}{sin^{2}(a)(1+cos^{2}(a))}+\frac{sin^{2}(a)}{cos^{2}(a)(1+sin^{2}(a))} \right ]cos^{2}(a)sin^{2}(a)
=\left [ \frac{cos^{2}(a)cos^{2}(a)(1+sin^{2}(a))+sin^{2}(a)sin^{2}(a)(1+cos^{2}(a))}{sin^{2}(a)(1+cos^{2}(a))cos^{2}(a)(1+sin^{2}(a))} \right ]cos^{2}(a)sin^{2}(a)
=\frac{cos^{4}(a)(1+sin^{2}(a))+sin^{4}(a)(1+cos^{2}(a))}{(1+cos^{2}(a))(1+sin^{2}(a))}
=\frac{cos^{4}(a)+cos^{4}(a)sin^{2}(a)+sin^{4}(a)+sin^{4}(a)cos^{2}(a))}{(1+cos^{2}(a)+sin^{2}(a)+cos^{2}(a)sin^{2}(a))}
=\frac{cos^{4}(a)+sin^{4}(a)+cos^{2}(a)sin^{2}(a)(sin^{2}(a)+cos^{2}(a))} {(1+1+cos^{2}(a)sin^{2}(a))}
=\frac{cos^{4}(a)+sin^{4}(a)+cos^{2}(a)sin^{2}(a)} {(2+cos^{2}(a)sin^{2}(a))}
=\frac{cos^{4}(a)+sin^{4}(a)+2cos^{2}(a)sin^{2}(a)-cos^{2}(a)sin^{2}(a)} {(2+cos^{2}(a)sin^{2}(a))}
=\frac{(cos^{2}(a)+sin^{2}(a))^{2}-cos^{2}(a)sin^{2}(a)} {(2+cos^{2}(a)sin^{2}(a))}
=\frac{1-cos^{2}(a)sin^{2}(a)} {2+cos^{2}(a)sin^{2}(a)}
=RHS
 
 
 
 
 
 
 
 
 
 
 
 
 
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Last Activity: 7 Years ago
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