If tan-125- find the value of 1-sin- 1 -sin -

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Relation between Trigonometric Ratios

If tanθ=125, ...

Question

If $\tan θ = \frac{12}{5}$ , find the value of $\frac{1 + \sin θ}{1 - \sin θ}$ .

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(i) $s e c θ (1 - s i n θ) (s e c θ + t a n θ) = 1$
(ii) $s i n θ (1 + t a n θ) + c o s θ (1 + c o t θ) = (s e c θ + c o s e c θ)$

sin θ (1 + tan θ) + c o s θ (1 + cot θ) = csc θ + sec θ

sec θ - tan θ = \frac{1}{4 x}

sec θ + tan θ

\frac{1}{sec θ - tan θ} + \frac{1}{sec θ + tan θ}

\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}

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