Question

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Answer 1

Answer:

here,

sinθ+cosθ=√3

orsquaring on both sides

(sinθ+cosθ)²=(√3)²

or, sin²θ+2sinθcosθ+cos²θ=3

or,1+2sinθcosθ=3

or,2sinθcosθ=2

or,sinθcosθ=1

or,sinθcosθ=sin²θ+cos²θ

or,1=(sin²θ+cos²θ)/(sinθcosθ)

or,1=[{sin²θ/(sinθcosθ)}+{cos²θ/(sinθcosθ)}]

or,1=[(sinθ/cosθ)+(cosθ/sinθ)]

1=tanθ+cotθ

∴tanθ+cotθ=1                      

                        PROVED

Answer 2

Answer:

sinx+cosx=√3

square both side

sin^2+cos^2+2sinxcosx=3

1+2sinxcox=3

sinxcosx=1

and prove tanx+cotx=1

so sinx/cosx+cosx/sinx=1

(sin^2x+cos^2)/sinxcosx=1

so. 1/sinxcos=1

we put the value of sinxcos=1

from above eq.

so 1/1=1

tanx+cotx=1