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sinθ/1+cosθ + 1+cosθ/sinθ is equal to

Question
$$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}= 2cosec θ$$
Solution
Let LHS be,
$$LHS =\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$$

Step 1: Take LCM of Denominator and Solve

$$\Rightarrow  LHS =\frac{\sin ^{2} \theta+(1+\cos \theta)^{2}}{\sin \theta(1+\cos \theta)}$$

$$\Rightarrow  LHS=\frac{\sin ^{2} \theta+1+2 \cos \theta+\cos ^{2} \theta}{\sin \theta(1+\cos \theta)}$$

Step 2: Rearrange the Equation
$$\Rightarrow  LHS=\frac{\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+1+2 \cos \theta}{\sin \theta(1+\cos \theta)}$$

Step 3: Put sin² θ +cos² θ= 1
$$\Rightarrow  LHS=\frac{1+1+2 \cos \theta}{\sin \theta(1+\cos \theta)}$$

$$\Rightarrow LHS=\frac{2+2 \cos \theta}{\sin \theta(1+\cos \theta)}$$

Step 4: Take out 2 common from the equation
$$=\frac{2(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$$

Step 5: Cancel out common terms
$$\Rightarrow  LHS=\frac{2}{\sin \theta}=2 \operatorname{cosec} \theta=RHS $$
LHS= RHS
Hence,
sinθ/1+cosθ + 1+cosθ/sinθ is equal to 2cosec θ