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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2026 · Mathematics · Trigonometric Ratios & Identities

If θ lies in the fourth quadrant and 3 θ+4=0 , then what is the...

If θ\theta lies in the fourth quadrant and 3cot⁡θ+4=03\cot\theta+4=0, then what is the value of sin⁡2θ+cos⁡2θ\sin 2\theta+\cos 2\theta ?

Options

  1. A−3125-\dfrac{31}{25}
  2. B−1725-\dfrac{17}{25}
  3. C00
  4. D11

Correct answer

B. −1725-\dfrac{17}{25}

Step-by-step solution

Given 3cot⁡θ+4=0⇒cot⁡θ=−433\cot\theta + 4 = 0 \Rightarrow \cot\theta = -\dfrac{4}{3}

Since θ\theta lies in the fourth quadrant, cos⁡θ>0\cos\theta > 0 and sin⁡θ<0\sin\theta < 0.

From cot⁡θ=−43\cot\theta = -\dfrac{4}{3}, we can consider the base as 44 and the perpendicular as 33.

Hypotenuse =42+32=5= \sqrt{4^2 + 3^2} = 5

Therefore, sin⁡θ=−35\sin\theta = -\dfrac{3}{5} and cos⁡θ=45\cos\theta = \dfrac{4}{5}.

Using double angle formulas:

sin⁡2θ=2sin⁡θcos⁡θ=2(−35)(45)=−2425\sin 2\theta = 2\sin\theta\cos\theta = 2\left(-\dfrac{3}{5}\right)\left(\dfrac{4}{5}\right) = -\dfrac{24}{25}

cos⁡2θ=cos⁡2θ−sin⁡2θ=(45)2−(−35)2=1625−925=725\cos 2\theta = \cos^2\theta - \sin^2\theta = \left(\dfrac{4}{5}\right)^2 - \left(-\dfrac{3}{5}\right)^2 = \dfrac{16}{25} - \dfrac{9}{25} = \dfrac{7}{25}

Adding both values:

sin⁡2θ+cos⁡2θ=−2425+725=−1725\sin 2\theta + \cos 2\theta = -\dfrac{24}{25} + \dfrac{7}{25} = -\dfrac{17}{25}

Answer: −1725-\dfrac{17}{25}
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