Question

If tan θ = 1, then find the value of `(sin θ + cos θ)/(sec θ + "cosec"  θ)` by completing the following activity.

Activity:

tan θ = 1

but tan `square` = 1

∴ θ = `square`

∴ `(sin θ + cos θ)/(sec θ + "cosec"  θ) = (sin 45^circ +  cos 45^circ)/(sec 45^circ +  "cosec"  45^circ)`

= `(1/square + 1/sqrt(2))/(sqrt(2)  +  square)`

= `(2/sqrt(2))/square`

`(sin θ + cos θ)/(sec θ + "cosec"  θ) = 1/square`

Answer 1

Activity:

tan θ = 1

but tan \[\boxed{45°}\] = 1

∴ θ = \[\boxed{45°}\]

∴ `(sin θ + cos θ)/(sec θ + "cosec"  θ) = (sin 45^circ +  cos 45^circ)/(sec 45^circ +  "cosec"  45^circ)`

= \[\frac{\frac{1}{\boxed{\sqrt2}} + \frac{1}{\sqrt2}}{\sqrt2 + \boxed{\sqrt2}}\]

= \[\frac{\frac{2}{\sqrt2}}{{\boxed{2\sqrt2}}}\]

`(sin θ + cos θ)/(sec θ + "cosec"  θ)` = \[\frac{1}{\boxed{2}}\]