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Step-by-step explanation:
here it is given tan A=1
so by expanding it
sinA/cosA=1
that implies
sinA=cosA
I.e., sinA-cosA=0
by squaring in both sides
(sinA-cosA) ^2=0
sinA^2+cosA^2-2sinAcosA=0
we know that
sinA^2+cosA^2=1
=(1)-2sinAcosA=0
=1=2sinAcosA
hence proved
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Given :
- In a Right triangle, Δ ABC right angled at B, tan A = 1
To prove :
- 2 sin A . cos A = 1
Proof :
Given that,
- tan A = 1
also, tan 45° = 1
so,
→ A = 45°
We need to prove that :
2 sin A . cos A = 1
taking LHS
→ LHS = 2 sin A . cos A
→ LHS = 2 sin 45° . cos 45°
putting value of sin 45° = 1 / √2 and cos 45° = 1 / √2
→ LHS = 2 ( 1 / √2 ) ( 1 / √2 )
→ LHS = 2 ( 1 / 2 )
→ LHS = 1 = RHS
PROVED .
Values of more trigonometric ratios :
- sin 0° = 0 ; sin 30° = 1 / 2 ; sin 45° = 1 / √2 ; sin 60° = √3 / 2 ; sin 90° = 1
- cos 0° = 1 ; cos 30° = √3 / 2 ; cos 45° = 1 / √2 ; cos 60° = 1 / 2 ; cos 90° = 0
- tan 0° = 0 ; tan 30° = 1 / √3 ; tan 45° = 1 ; tan 60° = √3 ; tan 90° = ∞ (undefined)
- cosec θ = 1 / sin θ
- sec θ = 1 / cos θ
- cot θ = 1 / tan θ
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