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CBSE Class 10 Math 2023 All Sets Solved Paper

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Question : 19 of 20
Marks: +1, -0
Statement A (Assertion): For 0<θ≤90∘,csc⁡θ−cot⁡θ0 < \theta \leq 90^{\circ}, \csc \theta - \cot \theta and csc⁡θ+cot⁡θ\csc \theta + \cot \theta are reciprocal of each other.
Statement R (Reason): cot⁡2θ−csc⁡2θ=1\cot^2 \theta - \csc^2 \theta = 1
Choose the correct option out of the following :
Solution:  
Statement R (Reason): cot⁡2θ−csc⁡2θ=1\cot^2 \theta - \csc^2 \theta = 1
Using trigonometric identity, ⇒1+cot⁡2θ=csc⁡2θ\Rightarrow 1 + \cot^2 \theta = \csc^2 \theta.
⇒cot⁡2θ−csc⁡2θ=−1\Rightarrow \cot^2 \theta - \csc^2 \theta = -1
Thus, Statement-2 is false.
Statement A (Assertion): For 0<θ≤90∘,csc⁡θ−cot⁡θ0 < \theta \leq 90^{\circ}, \csc \theta - \cot \theta and csc⁡θ+cot⁡θ\csc \theta + \cot \theta are reciprocal of each other.
csc⁡θ+cot⁡θ\csc \theta + \cot \theta
=(csc⁡θ+cot⁡θ)×csc⁡θ−cot⁡θcsc⁡θ−cot⁡θ=(\csc \theta + \cot \theta) \times \frac{\csc \theta - \cot \theta}{\csc \theta - \cot \theta}
=(csc⁡θ+cot⁡θ)(csc⁡θ−cot⁡θ)csc⁡θ−cot⁡θ= \frac{(\csc \theta + \cot \theta)(\csc \theta - \cot \theta)}{\csc \theta - \cot \theta}
=csc⁡2θ−cot⁡2θcsc⁡θ−cot⁡θ= \frac{\csc^2 \theta - \cot^2 \theta}{\csc \theta - \cot \theta}
=1csc⁡θ−cot⁡θ= \frac{1}{\csc \theta - \cot \theta} (∵1+cot⁡2θ=csc⁡2θ)(\because 1 + \cot^2 \theta = \csc^2 \theta)
For 0<θ≤90∘,csc⁡θ−cot⁡θ0 < \theta \leq 90^{\circ}, \csc \theta - \cot \theta and csc⁡θ+cot⁡θ\csc \theta + \cot \theta are reciprocal of each other.
Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.
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