Prove-
sin(90-theta).cos(90-theta) = tan theta / 1 + tan^2 theta
Answers
Answered by
0
Answer:
Step-by-step explanation:
LHS = sin(90°-theta) × cos(90°- theta)
= cos theta × sin theta
= sin theta × cos theta
RHS = tan theta / 1 + tan^2 theta
= tan theta / sec^2 theta
= (sin theta / cos theta) × cos^2 theta
= sin theta × cos theta
Answered by
1
Answer:
As we know that,
sin(90
o
−θ)=cosθ, cos(90
o
−θ)=sinθ,
cosec(90
o
−θ)=secθ, cot(90
o
−θ)=tanθ,
sec(90
o
−θ)=cosecθ, tan(90
o
−θ)=cotθ
∴
cosec(90−θ)sin(90−θ)cot(90−θ)
cos(90−θ)sec(90−θ)tanθ
+
cotθ
tan(90−θ)
∴
sin(90−θ)
1
×sin(90−θ)×tanθ
sinθ×
cos(90−θ)
1
×tanθ
+
cotθ
cotθ
=
cosθ
1
×cosθ
sinθ×
sinθ
1
+1
=1+1=2
Step-by-step explanation:
hope it's help you
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