Question

Express the followings terms into the asked ones.​

Answer 1

Answer:

1. Start with the ratio identity involving sine, cosine, and tangent, and multiply each side by cosine to get the sine alone on the left.

Replace cosine with its reciprocal function.

Solve the Pythagorean identity tan2θ + 1 = sec2θ for secant.

Replace the secant in the sine equation.

2. Sec thetha = 1/ cos theta

= 1sin/theta

Hence sec theta is 1/sin theta

Answer 2

Question :-

Express the following

$\rm :\longmapsto\:(i). \: \: sin\theta \: in \: terms \: of \: tan\theta$

$\rm :\longmapsto\:(ii). \: \: sec\theta \: in \: terms \: of \: sin\theta$

$\red{\large\underline{\sf{Solution-}}}$

Consider,

$\rm :\longmapsto\:sin\theta$

$\rm \:  =  \: \dfrac{1}{cosec\theta}$

$\rm \:  =  \: \dfrac{1}{ \sqrt{ {cosec}^{2}\theta } }$

$\rm \:  =  \: \dfrac{1}{ \sqrt{1 + {cot}^{2}\theta } }$

$\rm \:  =  \: \dfrac{1}{ \sqrt{1 + {\bigg(\dfrac{1}{tan\theta} \bigg) }^{2} } }$

$\rm \:  =  \: \dfrac{1}{ \sqrt{1 + \dfrac{1}{ {tan}^{2} \theta} } }$

$\rm \:  =  \: \dfrac{1}{ \sqrt{\dfrac{ {tan}^{2}\theta + 1}{ {tan}^{2} \theta} } }$

$\rm \:  =  \: \dfrac{tan\theta}{ \sqrt{ {tan}^{2} \theta + 1} }$

Hence,

$\red{\rm\implies \:\boxed{\tt{ \rm \: sin\theta =  \: \dfrac{tan\theta}{ \sqrt{ {tan}^{2} \theta + 1} } }}}$

$\green{\large\underline{\sf{Solution-ii}}}$

Consider

$\rm :\longmapsto\:sec\theta$

$\rm \:  =  \: \dfrac{1}{cos\theta}$

$\rm \:  =  \: \dfrac{1}{ \sqrt{ {cos}^{2} \theta} }$

$\rm \:  =  \: \dfrac{1}{ \sqrt{1 - {sin}^{2} \theta} }$

Thus,

$\green{\rm\implies \:\boxed{\tt{ \rm \: sec\theta =  \: \dfrac{1}{ \sqrt{1 - {sin}^{2} \theta} } }}}$

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Identities Used

$\boxed{\tt{ {sec}^{2}\theta - {tan}^{2}\theta = 1}}$

$\boxed{\tt{ {cosec}^{2}\theta - {cot}^{2}\theta = 1}}$

$\boxed{\tt{ {sin}^{2}\theta + {cos}^{2}\theta = 1}}$

$\boxed{\tt{ cot\theta = \frac{1}{tan\theta}}}$

$\boxed{\tt{ sec\theta = \frac{1}{cos\theta} }}$

$\boxed{\tt{ cosec\theta = \frac{1}{sin\theta} }}$