Question

prove: tanthita + 1/tanthita = secthita cosecthita​

Answer 1

$tan \theta + \frac{1}{tan \theta} \\ = \frac{ {tan}^{2} \theta + 1}{tan \theta} \\ = \frac{ {sec}^{2} \theta }{tan \theta} \\ = sec \theta \frac{sec \theta}{tan \theta} \\ = sec \theta \: cosec \theta$

Answer 2

To prove :-

${ \dashrightarrow\bigg( \tan \theta + \dfrac{1}{ \tan \theta} \bigg) = \sec \theta. cosec \theta}$

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Solution :-

${ \dashrightarrow\bigg( \tan \theta + \dfrac{1}{ \tan \theta} \bigg) = \sec \theta. cosec \theta}$

${ \dashrightarrow\bigg( \dfrac{ \tan^{2} \theta + 1}{ \tan \theta} \bigg) = \sec \theta. cosec \theta}$

» Apply formula tan²θ + 1 = sec²θ

${ \dashrightarrow\bigg( \dfrac{ \sec^{2} \theta }{ \tan \theta} \bigg) = \sec \theta. cosec \theta}$

» Apply formula tanθ = secθ/cosecθ

$\rm{ \dashrightarrow\Bigg( \dfrac{ \sec^{2} \theta }{ \frac{ \sec \theta}{ cosec \theta}} \Bigg) = \sec \theta. cosec \theta}$

$\rm{ \dashrightarrow\Bigg( \dfrac{ \sec^{2} \theta. cosec \theta }{ \sec \theta} \Bigg) = \sec \theta. cosec \theta}$

$\rm{ \dashrightarrow\bigg( \dfrac{ \ \cancel{sec^{2} \theta.} cosec \theta }{ \cancel{\sec \theta}} \bigg) = \sec \theta. cosec \theta}$

${ \dashrightarrow \boxed { \red{ \rm\sec \theta. cosec \theta = \sec \theta. cosec \theta}}}$

» LHS = RHS

Hence proved