Question

1+cot theta =cosec theta, then the general value of theta is​

Answer 1

Answer:

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Answer 2

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Step by step explanation:-

1+cot θ= cosecθ

1+cosθ/sinθ = 1/sinθ

Taking LCM in LHS side;

sinθ+cοsθ/sinθ =1/sinθ

sinθ( sinθ + cosθ)=sinθ

sin θ(sinθ + cosθ) -sinθ =0

sinθ[(sinθ + cosθ)-1]=0

Now,

Sinθ=0

(sinθ+cosθ-1)=0

We have two cases ;

Case I :- If sinθ=0,Τhen value of θ

sinθ =sin0°

sinθ =sin{nπ +(1)^n 0°}

[θ=nπ ]

Case II:-If sin θ+cosθ-1=0,then value of θ

sin θ+cosθ-1=0

sinθ + cosθ =1

squaring both side;

(sinθ + cosθ )^2 = (1)^2

(sin^2θ + cos^2θ +2sinθ cosθ)

Identity used here:-

sin^2θ + cos^2θ =1

So,

1+Sin^2θ =1

Sin^2θ =0°

Sin2θ = Sin{nπ +(-1)^n0°}

2θ = nπ

[θ=nπ/2]

(Hence we get the answer ;)

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