Question

If sin θ + cos θ = √2, then evaluate: tan θ + cot θ

Answer 1

Answer:

If sin θ + cos θ = √2, then evaluate: tan θ + cot θ

Step-by-step explanation:

sin θ + cos θ = √2

(sin θ + cos θ)2 = (√2)2

sin2θ + cos2θ + 2 sin θ cos θ = 2

1 + 2 sin θ cos θ = 2

sin θ cos θ = 1/2 ... (i)

We know that, sin2θ + cos2θ = 1 ... (ii)

Dividing (ii) by (i) we get

tan θ + cot θ = 2

Answer 2

Answer:

Squaring first equatio

Squaring first equatiosin^2(a)+cos^2(a) +2sin( a) cos( a)=2

Squaring first equatiosin^2(a)+cos^2(a) +2sin( a) cos( a)=21+2sin(a)cos(a)=2

Squaring first equatiosin^2(a)+cos^2(a) +2sin( a) cos( a)=21+2sin(a)cos(a)=22sin(a)cos(a)=1

Squaring first equatiosin^2(a)+cos^2(a) +2sin( a) cos( a)=21+2sin(a)cos(a)=22sin(a)cos(a)=1sin(a)cos(a)=1/2

Squaring first equatiosin^2(a)+cos^2(a) +2sin( a) cos( a)=21+2sin(a)cos(a)=22sin(a)cos(a)=1sin(a)cos(a)=1/2tan a+cot a=sin a/cos a +cos a/sin a

Squaring first equatiosin^2(a)+cos^2(a) +2sin( a) cos( a)=21+2sin(a)cos(a)=22sin(a)cos(a)=1sin(a)cos(a)=1/2tan a+cot a=sin a/cos a +cos a/sin a=sin ^2 a +cos ^2 a/sin a cos a

Squaring first equatiosin^2(a)+cos^2(a) +2sin( a) cos( a)=21+2sin(a)cos(a)=22sin(a)cos(a)=1sin(a)cos(a)=1/2tan a+cot a=sin a/cos a +cos a/sin a=sin ^2 a +cos ^2 a/sin a cos a=1/(1/2)

Squaring first equatiosin^2(a)+cos^2(a) +2sin( a) cos( a)=21+2sin(a)cos(a)=22sin(a)cos(a)=1sin(a)cos(a)=1/2tan a+cot a=sin a/cos a +cos a/sin a=sin ^2 a +cos ^2 a/sin a cos a=1/(1/2)=2