Question

) a) If sec A+tan A=P show that sin A = P2 -1
P2 +1​

Answer 1

Given :

secA + tanA = p

To prove :

sinA = (p² - 1)/(p² + 1)

Proof :

We have ;

secA + tanA = p --------(1)

Also ,

We know that , sec²A - tan²A = 1

Now ,

=> sec²A - tan²A = 1

=> (secA - tanA)·(secA + tanA) = 1

=> (secA - tanA)·p = 1

=> secA - tanA = 1/p --------(2)

Now ,

Subtracting eq-(2) and (1) , we get ;

=> (secA + tanA) - (secA - tanA) = p - 1/p

=> secA + tanA - secA + tanA = p - 1/p

=> 2tanA = (p² - 1)/p

=> tanA = (p² - 1)/2p ---------(3)

Now ,

Adding eq-(1) and (2) , we get ;

=> (secA + tanA) + (secA - tanA) = p + 1/p

=> 2secA = (p² + 1)/p

=> secA = (p² + 1)/2p ------(4)

Now ,

Dividing eq-(3) by (4) , we get ;

=> tanA/secA = [(p² - 1)/2p] / [(p² + 1)/2p]

=> tanA/secA = [(p² - 1)/2p] × [2p/(p² + 1)]

=> (sinA/cosA) / (1/cosA) = (p² - 1)/(p² + 1)

=> (sinA/cosA) × cosA = (p² - 1)/(p² + 1)

=> sinA = (p² - 1)/(p² + 1)

Hence proved .