Question

If cot theta is equal to root 7 then show that cosec squared theta minus sec square theta by cos square theta + sec square theta is equal to 3 by 4​

Answer 1

$Given ,\: cot\:theta = \sqrt{7} \: ---(1)$

$Value \: of \: \frac{cosec^{2} \theta - sec^{2} \theta }{ cosec^{2} \theta + sec^{2} \theta }$

/* Multiplying numerator and denominator by

$cos^{2} \theta$ , we get */

$= \frac{cos^{2} \theta(cosec^{2} \theta - sec^{2} \theta) }{cos^{2} \theta ( cosec^{2} \theta + sec^{2} \theta) }$

$= \frac{cos^{2} \theta\big( \frac{1}{sin^{2} \theta} - \frac{1}{cos^{2} \theta}\big)}{cos^{2} \theta \big(\frac{1}{sin^{2} \theta} + \frac{1}{cos^{2} \theta}\big)}$

$= \frac{ \big(\frac{ cos^{2} \theta}{sin^{2} \theta} - 1\big ) }{ \big(\frac{ cos^{2} \theta}{sin^{2} \theta} + 1\big )}\\= \frac{cot^{2} \theta - 1 }{cot^{2} \theta + 1 } \\= \frac{ \sqrt{7}^{2} - 1 }{\sqrt{7}^{2} + 1 }\:[From \:(1) ]$

$= \frac{ 7 - 1 }{ 7 + 1 }\\= \frac{6}{8} \\= \frac{3}{4}$

Therefore.,

$\red { Value \: of \: \frac{cosec^{2} \theta - sec^{2} \theta }{ cosec^{2} \theta + sec^{2} \theta }}\green { =\frac{3}{4}}$

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Answer 2

Step-by-step explanation:

Given ,\: cot\:theta = \sqrt{7} \: ---(1)Given,cottheta=

7

−−−(1)

Value \: of \: \frac{cosec^{2} \theta - sec^{2} \theta }{ cosec^{2} \theta + sec^{2} \theta }Valueof

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

/* Multiplying numerator and denominator by

cos^{2} \thetacos

2

θ , we get */

= \frac{cos^{2} \theta(cosec^{2} \theta - sec^{2} \theta) }{cos^{2} \theta ( cosec^{2} \theta + sec^{2} \theta) }=

cos

2

θ(cosec

2

θ+sec

2

θ)

cos

2

θ(cosec

2

θ−sec

2

θ)

= \frac{cos^{2} \theta\big( \frac{1}{sin^{2} \theta} - \frac{1}{cos^{2} \theta}\big)}{cos^{2} \theta \big(\frac{1}{sin^{2} \theta} + \frac{1}{cos^{2} \theta}\big)}=

cos

2

θ(

sin

2

θ

1

+

cos

2

θ

1

)

cos

2

θ(

sin

2

θ

1

−

cos

2

θ

1

)

\begin{gathered} = \frac{ \big(\frac{ cos^{2} \theta}{sin^{2} \theta} - 1) }{ \big(\frac{ cos^{2} \theta}{sin^{2} \theta} + 1)}\\= \frac{cot^{2} \theta - 1 }{cot^{2} \theta + 1 } \\= \frac{ \sqrt{7}^{2} - 1 }{\sqrt{7}^{2} + 1 }\:[From \:(1) ]\end{gathered}

=

(

sin

2

θ

cos

2

θ

+1)

(

sin

2

θ

cos

2

θ

−1)

=

cot

2

θ+1

cot

2

θ−1

=

7

2

+1

7

2

−1

[From(1)]

\begin{gathered} = \frac{ 7 - 1 }{ 7 + 1 }\\= \frac{6}{8} \\= \frac{3}{4} \end{gathered}

=

7+1

7−1

=

8

6

=

4

3

Therefore.,

\red { Value \: of \: \frac{cosec^{2} \theta - sec^{2} \theta }{ cosec^{2} \theta + sec^{2} \theta }}\green { =\frac{3}{4}}Valueof

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=

4

3

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