tanA+secA=P than prove that (p2+1)/(p2-1)=sinA
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p² - 1 = (sec a + tan a)² - 1
= sec² a + 2 sec a tan a + tan² a - 1
= 2 sec a tan a + (sec² a - 1) + tan² a
= 2 sec a tan a + 2 tan² a
= 2 tan a (sec a + tan a)
= 2p tan a.
Also,
p² + 1 = sec² a + 2 sec a tan a + tan² a + 1
= sec² a + 2 sec a tan a + (tan² a + 1)
= 2 sec² a + 2 sec a tan a
= 2 sec a(sec a + tan a)
= 2p sec a.
So,
(p² - 1) / (p² + 1) = (2p tan a) / (2p sec a)
= tan a / sec a
= tan a cos a
= sin a,
= sec² a + 2 sec a tan a + tan² a - 1
= 2 sec a tan a + (sec² a - 1) + tan² a
= 2 sec a tan a + 2 tan² a
= 2 tan a (sec a + tan a)
= 2p tan a.
Also,
p² + 1 = sec² a + 2 sec a tan a + tan² a + 1
= sec² a + 2 sec a tan a + (tan² a + 1)
= 2 sec² a + 2 sec a tan a
= 2 sec a(sec a + tan a)
= 2p sec a.
So,
(p² - 1) / (p² + 1) = (2p tan a) / (2p sec a)
= tan a / sec a
= tan a cos a
= sin a,
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