Question

given that cos theta=m/n and then tan theta is equal to.....
answer the this quickly please..... ​

Answer 1

GIVEN :-

• Cosθ = m/n

TO FIND :-

• Tanθ

SOLUTION :-

$\sf { \cos \theta = \frac{m}{n} = \dfrac{adjacent \: side}{hypotenuse} }$

Let us assume a rightangled ΔABC and θ is at C Then

• AB = m

• BC = n

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$\\ \sf (AC) {}^{2} = (AB) {}^{2} + (BC) {}^{2} \: [pythogoreas \: theorem] \\ \\ \implies \sf \: {n}^{2} = {m}^{2} + (BC) {}^{2} \\ \\ \implies \sf \: {n}^{2} - {m}^{2} = (BC) {}^{2} \\ \\ \implies {\bold {\boxed{\sf {\pink{(BC) = \sqrt{n {}^{2} - m {}^{2} } }}}}}$

$\sf \sin \theta = \frac{opposite \: side}{hypotenuse} = \frac{BC}{AC} = \frac{ \sqrt{ {n}^{2} - {m}^{2} } }{n} \\ \\ \sf \tan\theta = \frac{ \sin \theta}{ \cos\theta } = \frac{ \frac{ \sqrt{ {n}^{2} - {m}^{2} } }{ \cancel{n}} }{ \frac{m}{ \cancel{n}} } = \frac{ \sqrt{ {n}^{2} - m {}^{2} } }{m}$

$\therefore \sf \tan \theta = \frac{ \sqrt{ {n}^{2} - {m}^{2} } }{m}$

ADDITIONAL INFO :-

$(1) \sf \cosec\theta = \dfrac{1}{\sin \theta} \\ \\ (2) \sf \: \cot \theta = \frac{1}{ \tan \theta} \\ \\ (3) \sf \: \sec\theta = \frac{1}{ \cos \theta}$