1. sin A = cos (90o - A)
2.
3. tan A x tan (90o - A) = 1
4. sin2 A + cos2 A = 15. 1 + tan2 A = sec2 A
6. 1 + cot2 A = cosec2ALet us prove the above identities.
Let DABC be a right-angled triangle with ÐB = 90o.

…(1)

C = 90o - A
cos (90o - A)
…(2)
From (1) and (2),
sine of an angle = cosine of its complement.
Similarly we can prove that cos A = sin (90o - A)
and cos A



…(1)
also
…(2)
From (1) and (2),

…(1)

C = 90o - A
tan (90o - A)
…(2)
from (1) and (2),
tan A x tan (90o - A) = 1
We can use the above proof to show that tan (90o - A) = cot A
a2 + c2 = b2 (by Pythagoras Theorem)
by b2 
But sin
and 
(sin A)2 + (cos A)2 = 1
sin2A + cos2A = 1

by c2 

But
and 
(tan A)2 + 1 = (sec A)2
1 + tan2 A = sec2 A
Again a2 + c2 = b2 (by Pythagoras Theorem)
by a2


and cosec A
1 + (cot A)2 = (cosec A)2
1 + cot2 A = cosec2 A
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(
sin2 A + cos2 A = 1)
= 2 cosec2A = R.H.S.
Prove that 
L.H.S. 


= R.H.S.
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= cosec q - cot q = R.H.S.
Prove that 
L.H.S.

