Question

Tan²theta + cot ²theta +2 = sec²theta.cosec²theta

Answer 1

Step-by-step explanation:

tan²+cot²+2

tan²+cot²+2tancot

(tan+cot)²

=(1/sincos)²

=(sec cosec)²

=sec².cosec²

Answer 2

Solution →

${tan}^{2} \: \theta + {cot}^{2} \: \theta + 2 = {sec}^{2} \theta \times {cos}^{2} \: \theta$

LHS →

$= {tan}^{2} \: \theta + {cot}^{2} \: \theta + 2 (1)$

We know that →

• $tan \: \theta \times cot \: \theta = 1$

$= {tan}^{2} \: \theta + {cot}^{2} \: \theta + 2 \times tan \: \theta \: \times \: cot \: \theta$

We know that →

• a² + b² + 2ab = (a + b)²

Like this ,

$= {(tan \: \theta + cot \: \theta)}^{2}$

Now →

• $tan \: \theta \: = \frac{sin \: \theta}{cos \: \theta} \:$

$cot \: \theta = \frac{cos \: \theta}{sin \: \theta} \:$

So,

$= {( \frac{sin \: \theta}{cos \: \theta \: } + \frac{cos \: \theta}{sin \:\theta }) }^{2}$

$= {( \frac{ {sin}^{2}\theta + {cos \: \theta}^{2} }{sin \: \theta \times cos \: \theta}) }^{2}$

We know that →

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

So,

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

We know that →

• $\frac{1}{sin \: \theta} = cosec \: \theta \:$
• $\frac{1}{cos \: \theta} = sec \: \theta \:$

So,

$\boxed{\red{ = {cosec}^{2} \theta \times {sec}^{2} \theta}}$ RHS

Hence Proved!!

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