Prove that sec a {1-sin a} [sec a+ tana] = 1
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sec a{1-sin a} [sec a+ tan a]
(sec a- sec a sin a) [sec a+ tan a]
(sec a- 1/cos a sin a) [sec a+ tan a]
(sec a - sin a/cos a) [sec a+ tan a]
(sec a - tan a) [sec a+ tan a]
(sec^2 a - tan^2 a)
1
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