Prove that-cos-1-sin-tan-4-2

Cos θ/1 sin θ = cos^2 θ/2 sin^2 θ/2θ/sin^2 θ/2+cos^2 θ/2 2 sin θ/2.cos θ/2 [ ∵ cos 2A = cos^2 A sin^2 A; and sin^2 A + cos^2 A = 1 ] = ( cos θ/2 sin θ/2 ...

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Standard XII

Mathematics

Sin2A and Cos2A in Terms of tanA

Prove that:co...

Question

Prove that:

$\frac{c o s θ}{1 - s i n θ} = t a n (\frac{π}{4} + \frac{θ}{2})$

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Solution

$\frac{c o s θ}{1 - s i n θ} = \frac{c o s^{2} \frac{θ}{2} - s i n^{2} \frac{θ}{2} θ}{s i n^{2} \frac{θ}{2} + c o s^{2} \frac{θ}{2} - 2 s i n \frac{θ}{2} . c o s \frac{θ}{2}} [\begin{matrix} ∵ c o s 2 A = c o s^{2} A - s i n^{2} A and s i n^{2} A + c o s^{2} A = 1 \end{matrix}] = \frac{(c o s \frac{θ}{2} - s i n \frac{θ}{2}) (c o s \frac{θ}{2} + s i n \frac{θ}{2})}{(c o s \frac{θ}{2} - s i n \frac{θ}{2})^{2}} = \frac{c o s \frac{θ}{2} + s i n \frac{θ}{2}}{c o s \frac{θ}{2} - s i n \frac{θ}{2}}$

Dividing numerator and denominator by $c o s \frac{θ}{2}$

$= \frac{1 + t a n \frac{θ}{2}}{1 - t a n \frac{θ}{2}} = t a n (\frac{π}{4} + \frac{θ}{2}) = RHS$

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Prove that: cos θ1−sin θ=tan (π4+θ2)

cos θ1−sin θ=cos2 θ2−sin2 θ2θsin2 θ2+cos2 θ2−2 sin θ2.cos θ2[∵ cos 2A=cos2A−sin2 Aand sin2 A+cos2 A=1]=(cos θ2−sin θ2)(cos θ2+sin θ2)(cos θ2−sin θ2)2=cos θ2+sin θ2cos θ2−sin θ2 Dividing numerator and denominator by cos θ2 =1+tan θ21−tan θ2=tan (π4+θ2)=RHS

Prove that:

$\frac{c o s θ}{1 - s i n θ} = t a n (\frac{π}{4} + \frac{θ}{2})$