Question

Prove cot square theta minus tan square theta equal cosec square theta minus sec square theta

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

It has been proved that

$\bf \red{{cot}^{2} \: \theta - {tan}^{2} \: \theta = {cosec}^{2} \: \theta - {sec}^{2} \: \theta }$

Given:

• ${cot}^{2} \: \theta - {tan}^{2} \: \theta = {cosec}^{2} \: \theta - {sec}^{2} \: \theta$

To find:

• Prove the trigonometric equation.

Solution:

Identity to be used:

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

Step 1:

Rewrite the Identity.

Rewrite the Identity so as to use in LHS.

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

and

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

Step 2:

Apply Identity in LHS.

$= {cosec}^{2} \: \theta - 1 - ( {sec}^{2} \: \theta - 1)$

or

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

or

$= {cosec}^{2} \: \theta- {sec}^{2} \: \theta$

= RHS

Hence Proved.

Thus,

It has been proved that

$\bf {cot}^{2} \: \theta - {tan}^{2} \: \theta = {cosec}^{2} \: \theta - {sec}^{2} \: \theta$

Learn more:

1) Prove this identity in trigonometry

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

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