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If secA+tanA=p, then find the value of cosecA

Posted by Ankita Vijay 7 years, 1 month ago

CBSE > Class 10 > Mathematics

3 answers

Aamish Helal 7 years, 1 month ago

Arrange krlena ache se smjhke bnana

Aamish Helal 7 years, 1 month ago

secA+tanA=p ----------- (i) ‘.' sec²A-tan²A=1 (secA+tanA)(secA-tanA)=1 secA-tanA = 1/p ------------- (ii) Subtracting eqn ii from I We get, 2tanA= p-1/p tanA = (p²-1)/2p cotA = 2p/(p²-1) Now cosec²A-cot²A = 1 cosec²A=1+cot²A cosec²A=1+[2p/(p²-1)]² cosec²A= 1+4p²/(p²-1)² cosec²A= (p²)² - 2p²- 1- 4p²/(p²-1)² cosec²A= (p²)² + 2p²+1/(p²-1)² cosec²A= (p²+1)² / (p²-1)² cosecA= (p²+1)/p²-1)

Prantik Kalita 7 years, 1 month ago

secA+tanA=p ----------------------------(1) ∵, sec²A-tan²A=1 or, (secA+tanA)(secA-tanA)=1 or, secA-tanA=1/p -----------------------(2) Subtracting (2) from (1) we get, 2tanA=p-1/p or, tanA=(p²-1)/2p ∴, cotA=2p/(p²-1) Now, cosec²A-cot²A=1 or, cosec²A=1+cot²A or, cosec²A=1+{2p/(p²-1)}² or, cosec²A=1+4p²/(p²-1)² or, cosec²A=(p⁴-2p²+1+4p²)/(p²-1)² or, cosec²A=(p⁴+2p²+1)/(p²-1)² or, cosec²A=(p²+1)²/(p²-1)² or, cosecA=(p²+1)/(p²-1)

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