LoveUmyCrushidn
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Prove that :
following
[tex](tan \: q \: + sec \: q) {}^{2} = \frac{1 + sin \: q \: }{1 - sin \: q \: } [/tex]
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Sagot :

Loneguyidn

[tex]\text{L.H.S}\\\\=(\tan q + \sec q)^2\\\\\\=\left(\dfrac{\sin q}{ \cos q} + \dfrac 1{ \cos q} \right)^2\\\\\\=\left( \dfrac{1+ \sin q}{\cos q}\right)^2\\\\\\=\dfrac{(1+ \sin q)^2}{\cos^2 q}\\\\\\=\dfrac{(1+ \sin q)^2}{1-\sin^2 q}\\\\\\=\dfrac{(1 + \sin q)(1 + \sin q)}{(1+ \sin q)(1 - \sin q)}\\\\\\=\dfrac{1+ \sin q}{1- \sin q}\\\\=\text{R.H.S}\\\\\text{Proved.}[/tex]

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