Simplify the LHS of ( #10;sinθ #160; + θ)^2 #10;+ ( cosθ #160; + θ #10;)^2 = ( 1 + θθ)^2 #10; #10; ( sinθ #160; + θ)^2 + ( cos #10;θ ...

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Derivative of Standard Functions

sinθ+θ2+cosθ+...

Question

${(sin θ + sec θ)}^{2} + {(cos θ + csc θ)}^{2}$ $= {(1 + sec θ . csc θ)}^{2}$

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Solution

Simplify the LHS of ${(sin θ + sec θ)}^{2} + {(cos θ + csc θ)}^{2} = {(1 + sec θ csc θ)}^{2}$

${(sin θ + sec θ)}^{2} + {(cos θ + csc θ)}^{2} = {(sin θ + \frac{1}{cos θ})}^{2} + {(cos θ + \frac{1}{sin θ})}^{2}$

$= {(\frac{sin θ cos θ + 1}{cos θ})}^{2} + {(\frac{sin θ cos θ + 1}{sin θ})}^{2}$

$= {(1 + sin θ cos θ)}^{2} [\frac{1}{{cos}^{2} θ} + \frac{1}{{sin}^{2} θ}]$

$= \frac{{(1 + sin θ cos θ)}^{2}}{{sin}^{2} θ {cos}^{2} θ}$

$= {[\frac{(1 + sin θ cos θ)}{sin θ cos θ}]}^{2}$

$= {(1 + sec θ csc θ)}^{2}$

$= R H S$

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