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NCERT Exemplar · Q30

Q.If sin⁡θ+csc⁡θ=2\sin\theta + \csc\theta = 2, then sin⁡2θ+csc⁡2θ\sin^2\theta + \csc^2\theta is equal to
(A) 11
(B) 44
(C) 22
(D) None of these

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When a number and its reciprocal sum to 22, each must equal 11; squaring the original equation and adjusting for the cross-term reveals that sin⁡2θ+csc⁡2θ=2\sin^2\theta + \csc^2\theta = 2.

The heart of this problem lies in recognizing a special algebraic relationship. When sin⁡θ+csc⁡θ=2\sin\theta + \csc\theta = 2, we're being told that a number and its reciprocal add up to exactly 22. This is a very constrained situation—most pairs of reciprocals sum to something larger. The key insight is to square the given equation strategically, then isolate what we're looking for.

Let me denote x=sin⁡θx = \sin\theta for clarity. Then csc⁡θ=1x\csc\theta = \frac{1}{x}, and our condition becomes:

x+1x=2x + \frac{1}{x} = 2

Multiplying through by xx:

x2+1=2xx^2 + 1 = 2x

x2−2x+1=0x^2 - 2x + 1 = 0

(x−1)2=0(x - 1)^2 = 0

So x=1x = 1, meaning sin⁡θ=1\sin\theta = 1 and consequently csc⁡θ=1\csc\theta = 1.

Now I could substitute directly, but let me show the more general algebraic approach that works even when you don't immediately solve for the individual values.

Step-by-step derivation

  1. Square both sides of the given equation:

(sin⁡θ+csc⁡θ)2=22(\sin\theta + \csc\theta)^2 = 2^2

sin⁡2θ+2sin⁡θ⋅csc⁡θ+csc⁡2θ=4\sin^2\theta + 2\sin\theta \cdot \csc\theta + \csc^2\theta = 4

  1. Simplify the middle term:

    Notice that sin⁡θ⋅csc⁡θ=sin⁡θ⋅1sin⁡θ=1\sin\theta \cdot \csc\theta = \sin\theta \cdot \frac{1}{\sin\theta} = 1. This is the key simplification.

sin⁡2θ+2(1)+csc⁡2θ=4\sin^2\theta + 2(1) + \csc^2\theta = 4

  1. Isolate the desired expression: …

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