Prove the following:
cot2 A - 1sin2A\dfrac{1}{\text{sin}^2 A}sin2A1 + 1 = 0
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Solving L.H.S. of the equation : cot2 A - 1sin2A\dfrac{1}{\text{sin}^2 A}sin2A1 + 1 = 0
⇒cos2Asin2A−1sin2A+1⇒cos2A−1+sin2Asin2A⇒sin2A+cos2A−1sin2A⇒1−1sin2A⇒0sin2A⇒0.\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin}^2 A} - \dfrac{1}{\text{sin}^2 A} + 1 \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - 1 + \text{sin}^2 A}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A - 1}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{1 - 1}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{0}{\text{sin}^2 A} \\[1em] \Rightarrow 0.⇒sin2Acos2A−sin2A1+1⇒sin2Acos2A−1+sin2A⇒sin2Asin2A+cos2A−1⇒sin2A1−1⇒sin2A0⇒0.
Since, L.H.S. = R.H.S.
Hence, proved that cot2A−1sin2A+1=0\text{cot}^2 A - \dfrac{1}{\text{sin}^2 A} + 1 = 0cot2A−sin2A1+1=0
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11 + tan2A+11 + cot2A\dfrac{1}{\text{1 + tan}^2 A} + \dfrac{1}{\text{1 + cot}^2 A}1 + tan2A1+1 + cot2A1 = 1
Simplify 1−sin2 θ1−cos2 θ\sqrt{\dfrac{1 - \text{sin}^2 \text{ θ}}{1 - \text{cos}^2 \text{ θ}}}1−cos2 θ1−sin2 θ.
Using the measurements given in the figure alongside,
(a) Find the values of:
(i) sin Φ
(ii) tan θ.
(b) Write an expression for AD in terms of θ.
(sin A + cos A)2 + (sin A - cos A)2 = 2
