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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2021 · Mathematics · Trigonometric Ratios & Identities

If 7 θ+24 θ=25 , then what is the value of ( θ+ θ) ?

If 7sin⁡θ+24cos⁡θ=257 \sin \theta+24 \cos \theta=25, then what is the value of (sin⁡θ+cos⁡θ)(\sin \theta+\cos \theta) ?

Options

  1. A11
  2. B2625\frac{26}{25}
  3. C65\frac{6}{5}
  4. D3125\frac{31}{25}

Correct answer

D. 3125\frac{31}{25}

Step-by-step solution

7sin⁡θ+24cos⁡θ=25⇒725⋅sin⁡θ+2425⋅cos⁡θ=1Again, (7)2+(24)2=(25)2\begin{aligned} & 7 \sin \theta+24 \cos \theta=25 \\ & \Rightarrow \frac{7}{25} \cdot \sin \theta+\frac{24}{25} \cdot \cos \theta=1 \\ & \text{Again, }(7)^2+(24)^2=(25)^2\end{aligned}\thereforeLetLet\frac{7}{25}=\cos \alphaandand\frac{24}{25}=\sin \alphaThen,Then,\sin \theta \cdot \cos \alpha+\cos \theta \cdot \sin \alpha=1⇒sin⁡(θ+α)=sin⁡π2⇒θ+α=π2⇒θ=(π2−α)\begin{aligned} & \Rightarrow \sin (\theta+\alpha)=\sin \frac{\pi}{2} \\ & \Rightarrow \theta+\alpha=\frac{\pi}{2} \Rightarrow \theta=\left(\frac{\pi}{2}-\alpha\right)\end{aligned}Now,Now,\sin \theta+\cos \theta=\sin (π2−α)+cos⁡(π2−α)\left(\frac{\pi}{2}-\alpha\right)+\cos \left(\frac{\pi}{2}-\alpha\right)
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