Math, asked by sonusharma45, 9 months ago

solve this mate plzz ​

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Answered by raghuannamalai1975
4

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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Answered by Cosmique
22

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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