Question

(1+tan theta)^2 +(1+cot theta)^2=(sec theta +cosec theta)^2 prove it

Answer 1

hope this helps..........

Answer 2

$\bf \red{( {1 + tan \: \theta })^{2} + ( {1 + cot \: \theta })^{2} = ( {sec \: \theta + cosec \: \theta })^{2} }$

The trigonometric equation has been proved.

Given:

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

To find:

• Prove the trigonometric equation.

Solution:

Identities to be used:

1. $( {x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}$
2. $cos^{2}\: \theta + {sin}^{2} \theta=1$
3. $1+tan^2\: \theta = {sec}^{2} \theta$
4. $1+cot^{2}\: \theta = {cosec}^{2} \theta$
5. $\frac{1 }{{sin} \: \theta}={cosec} \theta$
6. $\frac{1 }{{cos} \: \theta}={sec} \theta$
7. $\frac{cos\: \theta }{{sin} \: \theta}={cot} \theta$
8. $\frac{sin\: \theta }{{cos} \: \theta}={tan} \:\theta$

Step 1:

Open Identity 1 in LHS.

$\frac{1 }{{sin} \: \theta}={cosec} \theta$

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

or

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

Step 2:

Apply trigonometric Identity 3 and 4.

$= 1 + {sec}^{2} \theta - 1+ 2{tan} \theta + 1 + {cosec}^{2} \theta - 1 + 2{cot} \theta$

or

$= {sec}^{2} \theta+ 2{tan} \theta + {cosec}^{2} \theta + 2{cot} \theta$

or

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

Step 3:

Apply Identity 7 and 8 in bracket.

$= {sec}^{2} \theta + {cosec}^{2} \theta + 2 \left( \frac{{cos} \: \theta }{{sin} \: \theta } + \frac{{sin} \: \theta }{{cos} \: \theta } \right)$

or

$= {sec}^{2} \theta + {cosec}^{2} \theta + 2 \left( \frac{{cos^{2} } \: \theta + {sin}^{2} \theta }{{sin} \: \theta \: cos \: \theta} \right)$

or

Apply Identity 2.

$= {sec}^{2} \theta + {cosec}^{2} \theta + 2 \left( \frac{1 }{{sin} \: \theta \: cos \: \theta} \right)$

or

$= {sec}^{2} \theta + {cosec}^{2} \theta + 2 \frac{1 }{{sin} \: \theta}. \frac{1 }{{cos} \: \theta}$

or

Apply Identity 5 and 6.

$= {sec}^{2} \theta + {cosec}^{2} \theta + 2 cosec \: \theta. {sec} \: \theta$

or

Apply Identity 1.

$= ( {sec \: \theta +cosec \: \theta})^{2}$

= RHS

Thus,

The trigonometric equation has been proved.

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

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