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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2017 · Mathematics · Trigonometric Ratios & Identities

If θ-cosec θ=(4)/(3) , then what is ( θ- θ) equal to?

If sec⁡θ−cosec⁡θ=43\sec \theta-\operatorname{cosec} \theta=\frac{4}{3}, then what is (sin⁡θ−cos⁡θ)(\sin \theta-\cos \theta) equal to?

Options

  1. A-2 only
  2. B12\frac{1}{2} only
  3. CBoth -2 and \frac{1}{2}
  4. DNeither \frac{1}{2}$ nor -2

Correct answer

B. 12\frac{1}{2} only

Step-by-step solution

sec⁡θ−cosec⁡θ=43⇒1cos⁡θ−1sin⁡θ=43⇒sin⁡θ−cos⁡θsin⁡θcos⁡θ=43⇒(sin⁡θ−cos⁡θ)2(sin⁡θcos⁡θ)2=169⇒sin⁡2θ+cos⁡2θ−2sin⁡θcos⁡θsin⁡2θcos⁡2θ=169⇒1−2sin⁡θcos⁡θsin⁡2θcos⁡2θ=169\begin{aligned} & \sec \theta-\operatorname{cosec} \theta=\frac{4}{3} \Rightarrow \frac{1}{\cos \theta}-\frac{1}{\sin \theta}=\frac{4}{3} \\ & \Rightarrow \frac{\sin \theta-\cos \theta}{\sin \theta \cos \theta}=\frac{4}{3} \\ & \Rightarrow \frac{(\sin \theta-\cos \theta)^2}{(\sin \theta \cos \theta)^2}=\frac{16}{9} \\ & \Rightarrow \frac{\sin ^2 \theta+\cos ^2 \theta-2 \sin \theta \cos \theta}{\sin ^2 \theta \cos ^2 \theta}=\frac{16}{9} \\ & \Rightarrow \frac{1-2 \sin \theta \cos \theta}{\sin ^2 \theta \cos ^2 \theta}=\frac{16}{9} \end{aligned} Let sin⁡θcos⁡θ=x⇒1−2xx2=169⇒16x2+18x−9=0⇒(8x−3)(2x+3)=0⇒x=38,x=−32∴sin⁡θcos⁡θ=38\begin{aligned} & \text{ Let } \sin \theta \cos \theta=x \Rightarrow \frac{1-2 x}{x^2}=\frac{16}{9} \\ & \Rightarrow 16 x^2+18 x-9=0 \\ & \Rightarrow(8 x-3)(2 x+3)=0 \\ & \Rightarrow x=\frac{3}{8}, x=\frac{-3}{2} \\ & \therefore \sin \theta \cos \theta=\frac{3}{8} \end{aligned}from(i),from (i),\sin \theta-\cos \theta=\frac{4}{3} \times \frac{3}{8}=\frac{4}{8}=\frac{1}{2}$
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