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The value of  50+50cot2A25+25tan2A\frac{50 + 50 cot^2 A}{25 + 25 tan^2 A}25+25tan2A50+50cot2A​​  is:
Question

The value of  50+50cot2A25+25tan2A\frac{50 + 50 cot^2 A}{25 + 25 tan^2 A}​  is:

A.

2 cot2^2 A​

B.

12\frac{1}{2} cot2^2 A​

C.

2 tan2^2​ A

D.

12\frac{1}{2}​ tan2^2​ A

Correct option is A

Given:

 50+50cot2A25+25tan2A\frac{50 + 50 cot^2 A}{25 + 25 tan^2 A}   

Identity Used: 

1+cot2A=Cosec2A1+tan2A=Sec2A1 + cot^2A = Cosec^2 A \\1 + tan^2 A = Sec^2 A  

Solution: 

50+50cot2A25+25tan2A\frac{50 + 50 cot^2 A}{25 + 25 tan^2 A} 

=50+50cot2A25+25tan2A =50(1+cot2A)25(1+tan2A) =2×cosec2Asec2A =2×cos2Asin2A =2Cot2A= \frac{50 + 50 cot^2 A}{25 + 25 tan^2 A} \\\ \\= \frac{50(1 + cot^2 A)}{25 (1 + tan^2 A)} \\\ \\= \frac{2\times cosec^2A } {sec^2 A} \\\ \\= \frac{ 2 \times cos^2A}{sin^2A} \\\ \\= 2Cot^2A

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