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
sin2 θ
sec2 θ
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Given,
cot2 θ−1sin2 θ\text{cot}^2 \text{ θ} - \dfrac{1}{\text{sin}^2 \text{ θ}}cot2 θ−sin2 θ1
The equation can be written as,
⇒cos2 θsin2 θ−1sin2 θ⇒cos2 θ−1sin2 θ⇒−(1−cos2 θ)sin2 θ⇒−sin2 θsin2 θ⇒−1.\Rightarrow \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}} - \dfrac{1}{\text{sin}^2 \text{ θ}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 \text{ θ} - 1}{\text{sin}^2 \text{ θ}} \\[1em] \Rightarrow \dfrac{-(1 - \text{cos}^2 \text{ θ})}{\text{sin}^2 \text{ θ}} \\[1em] \Rightarrow \dfrac{-\text{sin}^2 \text{ θ}}{\text{sin}^2 \text{ θ}} \\[1em] \Rightarrow -1.⇒sin2 θcos2 θ−sin2 θ1⇒sin2 θcos2 θ−1⇒sin2 θ−(1−cos2 θ)⇒sin2 θ−sin2 θ⇒−1.
Hence, Option 2 is the correct option.
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If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 - y2 = a2 - b2.
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(sec2 θ - 1)(1 - cosec2 θ) is equal to
0
2
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
cos2 θ