Answer 1

`tan theta = 1/sqrt7`               `(cosec^2 theta - sec^2 theta)/(cosec^2 theta + sec^2 theta) = 3/4`

`tan theta = "𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒"/"𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒"`

Let ‘x’ be the hypotenuse

By applying Pythagoras

𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²

`x^2 = 1^2 + (sqrt7)^2`

𝑥² = 1 + 7 = 8

`x = 2sqrt2`

`cosec theta = (AC)/(AB) = 2sqrt2`

`sec theta = (AC)/(BC) = (2sqrt2)/sqrt7`

Substitute, cosec θ, sec θ in equation

`=> ((2sqrt2)^2 - (2 sqrt(2/7))^2)/((2sqrt2)^2 + ((2sqrt2)/sqrt7)^2)`

`(8 - 4 xx 2/7)/(8 + 4 xx 2/7)`

`=> (8 - 8/7)/(8 + 8/7)`

`=> ((56 - 8)/7)/((56 + 8)/7)`

`=48/64`

`= 3/4`

L.H.S = R.H.S