Question

tan theta+1/tan theta=
plz tell its urgent​

Answer 1

Answer:

LHS = tan\theta+\frac{1}{tan\theta}tanθ+

tanθ

1

/* LCM = tan\thetatanθ */

= \frac{(tan^{2}\theta+1)}{tan\theta}

tanθ

(tan

2

θ+1)

= \frac{sec^{2}\theta}{tan\theta}

tanθ

sec

2

θ

/* By Trigonometric identity:*/

\boxed{1+tan^{2}\theta = sec^{2}\theta}

1+tan

2

θ=sec

2

θ

= \frac{\frac{1}{cos^{2}\theta}}{\frac{sin\theta}{cos\theta}}

cosθ

sinθ

cos

2

θ

1

= \frac{1}{sin\theta cos\theta}

sinθcosθ

1

= \frac{1}{sin\theta}\times\frac{1}{cos\theta}

sinθ

1

×

cosθ

1

= cosec\theta sec\thetacosecθsecθ

________________________

Since ,

i) \frac{1}{sin\theta} = cosec\theta

sinθ

1

=cosecθ

ii)\frac{1}{cos\theta} = sec\theta

cosθ

1

=secθ

Answer 2

$\tan(θ) + \frac{1}{ \tan(θ)}$

$\frac{ \tan^{2} (θ) + 1}{ \tan(θ) }$

$\frac{ \sec^{2} (θ) }{ \tan(θ) }$

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