tan2 θ1 + tan2 θ\dfrac{\text{tan}^2 \text{ θ}}{\text{1 + tan}^2 \text{ θ}}1 + tan2 θtan2 θ is equal to
2sin2 θ
2cos2 θ
sin2 θ
cos2 θ
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Given, tan2 θ1 + tan2 θ\dfrac{\text{tan}^2 \text{ θ}}{\text{1 + tan}^2 \text{ θ}}1 + tan2 θtan2 θ
On solving,
⇒sin2 θcos2 θ1+sin2 θcos2 θ⇒sin2 θcos2 θcos2 θ+sin2 θcos2 θ⇒sin2 θ cos2 θcos2 θ(cos2 θ+sin2 θ)⇒sin2 θ.\Rightarrow \dfrac{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{1 + \dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{\dfrac{\text{cos}^2 \text{ θ} + \text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 \text{ θ} \text{ cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ} (\text{cos}^2 \text{ θ} + \text{sin}^2 \text{ θ})} \\[1em] \Rightarrow \text{sin}^2 \text{ θ}.⇒1+cos2 θsin2 θcos2 θsin2 θ⇒cos2 θcos2 θ+sin2 θcos2 θsin2 θ⇒cos2 θ(cos2 θ+sin2 θ)sin2 θ cos2 θ⇒sin2 θ.
Hence, Option 3 is the correct option.
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cot2 θ−1sin2 θ\text{cot}^2 \text{ θ} - \dfrac{1}{\text{sin}^2 \text{ θ}}cot2 θ−sin2 θ1 is equal to
1
-1
sec2 θ
(sec2 θ - 1)(1 - cosec2 θ) is equal to
0
2
(cos θ + sin θ)2 + (cos θ - sin θ)2 is equal to
-2
(sec A + tan A)(1 - sin A) is equal to
sec A
sin A
cosec A
cos A
