Prove- 1-cos2-cos-1-tan2-tan- - tan-

(1 cos^2θ)cosθ(1+tan^2θ)tanθ = tanθθ sin^2θcosθ^2 θtanθ = tanθθ tanθθ = tanθθ ...

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Circular Measurement of Angle

Prove: 1-co...

Question

Prove: $\frac{(1 - {cos}^{2} θ)}{cos θ} \frac{(1 + {tan}^{2} θ)}{tan θ} = tan θ sec θ$

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\frac{\sin^{2} θ}{cosθ} + cosθ = secθ

\cos^{2} θ (1 + \tan^{2} θ) = 1

\sqrt{\frac{1 - sinθ}{1 + sinθ}} = secθ - tanθ

(secθ - cosθ) (cotθ + tanθ) = tanθ secθ

cotθ + tanθ = cosecθ secθ

\frac{1}{secθ - tanθ} = secθ + tanθ

\sec^{4} θ - \cos^{4} θ = 1 - 2 \cos^{2} θ

secθ + tanθ = \frac{\cos θ}{1 - sinθ}

t anθ + \frac{1}{tanθ} = 2

\tan^{2} θ + \frac{1}{\tan^{2} θ} = 2

\frac{tanA}{{(1 + \tan^{2} A)}^{2}} + \frac{cotA}{{(1 + \cot^{2} A)}^{2}} = \sin A \cos A

\sec^{4} A (1 - \sin^{4} A) - 2 \tan^{2} A = 1

\frac{tanθ}{secθ - 1} = \frac{tanθ + secθ + 1}{tanθ + secθ - 1}

tan θ + \frac{1}{tan θ} = 2

{tan}^{2} θ + \frac{1}{{tan}^{2} θ} = 2

\frac{s i n 2 θ}{1 - c o s 2 θ} = c o t θ o r \frac{1 - c o s^{2} θ}{1 + c o s 2 θ} = \frac{t a n^{2} θ}{2}

t a n^{2} θ - \frac{1}{c o s^{2} θ} = - 1

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