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अगर 1+cos2θ=3sinθcosθ1+\cos ^{2} \theta=3 \sin \theta \cos \theta, तो cotθ(0<θ<π2काअभिन्नमूल्य)\cot \theta\left(0<\theta<\frac{\pi}{2} का अभिन्न मूल्य \right) is

a

1

b

2

c

0

d

3

Answer : Option A
Explanation :
1 + cos²θ = 3 sinθ . cosθ दोनों पक्षों को विभाजित करके sin²θ 1sin2θ+cos2θsin2θ=3sinθcosθsin2θ\frac{1}{\sin ^{2} \theta}+\frac{\cos ^{2} \theta}{\sin ^{2} \theta}=\frac{3 \sin \theta \cos \theta}{\sin ^{2} \theta} cosec2θ+cot2θ=3cotθ\Rightarrow \operatorname{cosec}^{2} \theta+\cot ^{2} \theta=3 \cot \theta 1+cot2θ+cot2θ=3cotθ\Rightarrow 1+\cot ^{2} \theta+\cot ^{2} \theta=3 \cot \theta 2cot2θ3cotθ+1=0\Rightarrow 2 \cot ^{2} \theta-3 \cot \theta+1=0 2cot2θ2cotθcotθ+1=0\Rightarrow 2 \cot ^{2} \theta-2 \cot \theta-\cot \theta+1=0 2cot2θ(cotθ1)1(cotθ1)=0\Rightarrow 2 \cot ^{2} \theta(\cot \theta-1)-1(\cot \theta-1)=0 (2cotθ1)(cotθ1)=0\Rightarrow(2 \cot \theta-1)(\cot \theta-1)=0 cotθ=12\Rightarrow \cot \theta=\frac{1}{2} or 1

विकल्प:

Let θ=45\theta=45^{\circ}

1+cos2θ=3sinθcosθ\Rightarrow 1+\cos ^{2} \theta=3 \sin \theta \cos \theta

1+(12)2=3×12×121+\left(\frac{1}{\sqrt{2}}\right)^{2}=3 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}

32=32\frac{3}{2}=\frac{3}{2} \quad (Satisfies)

Now, cotθ=cot45=1\cot \theta=\cot 45^{\prime}=1

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