If tan 20- p- then tan 160-tan 110-1-tan 160-tan 110-2 p-1-p21-p2-2 pC- 1-p2-1-p2D- 2 p-1-p2

The correct option is B G.E = tan (160^∘ 110^∘) = tan 50^∘ = 40^∘ = 1 tan^2 20^∘/2 tan 20^∘ = 1 p^2/2p. ...

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Fundamental Laws of Logarithms

If tan 20∘= p...

Question

If $tan 20^{\circ} = p, t h e n \frac{tan 160^{\circ} - tan 110^{\circ}}{1 + tan 160^{\circ} tan 110^{\circ}} =$

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tan 20^{\circ} = λ

\frac{tan 160^{\circ} - tan 110^{\circ}}{1 + tan 160^{\circ} \cdot tan 110^{\circ}} = \frac{1 - λ^{2}}{2 λ}

tan 20^{\circ} = k,

\frac{tan 160^{\circ} - tan 110^{\circ}}{1 + tan 160^{\circ} tan 110^{\circ}} = \frac{1 - k^{2}}{2 k}

if cot 20=p then [tan 160-tan110/]/1+tan 160*tan 110

\frac{tan 160^{\circ} - tan 110^{\circ}}{1 + tan 160^{\circ} . tan 110^{\circ}}

If $s e c θ + t a n θ = p$ , prove that

(i) $s e c θ = \frac{1}{2} (p + \frac{1}{p})$

(ii) $t a n θ = \frac{1}{2} (p - \frac{1}{p})$

(iii) $s i n θ = \frac{p^{2} - 1}{p^{2} + 1}$

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