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secondary 3 | A Maths
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pls help!
Date Posted: 4 years ago
Views: 368
(a)
tan² x + 1 = sec² x
so
1 - tan² x = 1 - (sec² x - 1)
= 2 - sec² x
&
1 + tan² x = 1 + (sec² x - 1)
= sec² x
thus
(1 - tan² x )/(1 + tan² x )
= (2 - sec² x)/sec² x
= 2/sec² x - 1
= 2cos² x - 1
.
tan² x + 1 = sec² x
so
1 - tan² x = 1 - (sec² x - 1)
= 2 - sec² x
&
1 + tan² x = 1 + (sec² x - 1)
= sec² x
thus
(1 - tan² x )/(1 + tan² x )
= (2 - sec² x)/sec² x
= 2/sec² x - 1
= 2cos² x - 1
.
See 1 Answer
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We simply use our few basic identities and techniques such as combining into single fraction, factorising into brackets, expanding brackets, introducing conjugate etc to prove these more complicated identities.
Date Posted:
4 years ago
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