Prove the following :
sin2 θcos θ+cos θ=1cos θ\dfrac{\text{sin}^2 \text{ θ}}{\text{cos \text{ θ}}} + \text{cos \text{ θ}} = \dfrac{1}{\text{cos \text{ θ}}}cos θsin2 θ+cos θ=cos θ1
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To prove,
Solving LHS of the above equation,
=sin2 θcos θ+cos θ=sin2 θ+cos2θcos θ=1cos θ [∵sin2 θ+ cos2 θ=1]\phantom{=} \dfrac{\text{sin}^2 \text{ θ}}{\text{cos θ}} + \text{cos θ} \\[1em] = \dfrac{\text{sin}^2 \text{ θ} + \text{cos}^2 θ}{\text{cos θ}} \\[1em] = \dfrac{1}{\text{cos θ}} \space [\because \text{sin}^2 \text{ θ} + \text{ cos}^2 \text{ θ} = 1]=cos θsin2 θ+cos θ=cos θsin2 θ+cos2θ=cos θ1 [∵sin2 θ+ cos2 θ=1]
Since, L.H.S. = R.H.S.
Hence, proved that sin2 θcos θ+cos θ=1cos θ.\dfrac{\text{sin}^2 \text{ θ}}{\text{cos θ}} + \text{cos θ} = \dfrac{1}{\text{cos θ}}.cos θsin2 θ+cos θ=cos θ1.
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cos θ tan θ = sin θ
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