Question

if tan theta equal to 1 by root 7 find the value of cosec squared theta minus sin square theta by cos square theta + sec square theta ​

Answer 1

answer : 441/113

given, $tan\theta=\frac{1}{\sqrt{7}}=\frac{p}{b}$

so, p = 1 and b = √7

from Pythagoras theorem,

h² = p² + b² => h = √(1² + √7²) = 2√2

hence, $cos\theta=\frac{b}{h}=\frac{\sqrt{7}}{2\sqrt{2}}$

so, $sec\theta=\frac{2\sqrt{2}}{\sqrt{7}}$

$sin\theta=\frac{p}{h}=\frac{1}{2\sqrt{2}}$

so, $cosec\theta=\frac{2\sqrt{2}}{1}$

we have to find $\frac{cosec^2\theta-sin^2\theta}{cos^2\theta+sec^2\theta}$

= $\frac{(2\sqrt{2})^2-\left(\frac{1}{2\sqrt{2}}\right)^2}{\left(\frac{\sqrt{7}}{2\sqrt{2}}\right)^2+\left(\frac{2\sqrt{2}}{\sqrt{7}}\right)^2}$

= $\frac{8-\frac{1}{8}}{\frac{7}{8}+\frac{8}{7}}$

= $\frac{64-1}{8\frac{(49+64)}{56}}$

= $\frac{63\times7}{113}$

= $\frac{441}{113}$

Answer 2

Step-by-step explanation:

answer : 441/113

given, tan\theta=\frac{1}{\sqrt{7}}=\frac{p}{b}tanθ=

7

1

=

b

p

so, p = 1 and b = √7

from Pythagoras theorem,

h² = p² + b² => h = √(1² + √7²) = 2√2

hence, cos\theta=\frac{b}{h}=\frac{\sqrt{7}}{2\sqrt{2}}cosθ=

h

b

=

2

2

7

so, sec\theta=\frac{2\sqrt{2}}{\sqrt{7}}secθ=

7

2

2

sin\theta=\frac{p}{h}=\frac{1}{2\sqrt{2}}sinθ=

h

p

=

2

2

1

so, cosec\theta=\frac{2\sqrt{2}}{1}cosecθ=

1

2

2

we have to find \frac{cosec^2\theta-sin^2\theta}{cos^2\theta+sec^2\theta}

cos

2

θ+sec

2

θ

cosec

2

θ−sin

2

θ

= \frac{(2\sqrt{2})^2-\left(\frac{1}{2\sqrt{2}}\right)^2}{\left(\frac{\sqrt{7}}{2\sqrt{2}}\right)^2+\left(\frac{2\sqrt{2}}{\sqrt{7}}\right)^2}

(

2

2

7

)

2

+(

7

2

2

)

2

(2

2

)

2

−(

2

2

1

)

2

= \frac{8-\frac{1}{8}}{\frac{7}{8}+\frac{8}{7}}

8

7

+

7

8

8−

8

1

= \frac{64-1}{8\frac{(49+64)}{56}}

8

56

(49+64)

64−1

= \frac{63\times7}{113}

113

63×7

= \frac{441}{113}

113

441