Question

please answer this one

Answer 1

Option D is correct

theta is written as @

if tan@ = 1/root7 = height/base 

then, let height = 1x and base = root7

then, by pythagoras theorem,

hypotenuse^2 = 1^2 +root7^2

hypotenuse^2 = 1+7

hypotenuse = root 8 = 2root2

then, cosec^2@ =
(hypotenuse/height)^2=8/1

and sec^2@ = (hypotenuse/base)^2 = 8/7

then,

cosec^2@ - sec^2@ = 8/1 -8/7 = (56-8)/7 =48/7 

and 

cosec^2@  + sec^2@ =8/1+8/7 = (56+8)/7 = 64/7

now,

(48/7)/(64/7)

=(48/64)

=24/32

=12/16

=3/4

i hope this will help 

Answer 2

Answer:

Option (D) - 3/4

Explanation:

Given tan theta = 1/root 7.

We know that sec^2theta = 1 + tan^2theta

                                           = 1 + (1/root 7)^2

                                           = 1 + 1/7

                                           = 8/7.    ----------------------- (1)


Now,

We know that cot theta = 1/tan theta

                                        = 1/1/root 7

                                        = root 7/1.



We know that cosec^2 theta = 1 + cot^2 theta

                                                 = 1 + (root 7/1)^2

                                                 = 1 + (7/1)

                                                 = 8/1.


Cosec theta = root 8/1.   ------------------------------ (2)


Given:

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

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

$\frac{ \frac{56 - 8}{7} }{ \frac{56 + 8}{7} }$

$\frac{48}{7} * \frac{7}{64}$

$\frac{3}{4}$


Hope this helps!