If
secθ−tanθ
secθ+tanθ
​
=
3
5
​
, 0
∘
<θ<90
∘
, then what is the value of (cscθ+cosθ+cotθ)?

Given:
secθ−tanθ
secθ+tanθ
​
=
3
5
​

Applying Componendo and Dividendo:

(secθ+tanθ)−(secθ−tanθ)
(secθ+tanθ)+(secθ−tanθ)
​
=
5−3
5+3
​

2tanθ
2secθ
​
=
2
8
​

sinθ/cosθ
1/cosθ
​
=4⟹
sinθ
1
​
=4⟹sinθ=
4
1
​

From sinθ=
4
1
​
:

cscθ=4
cosθ=
1−sin
2
θ

​
=
1−
16
1
​


​
=
4
15

​

​

cotθ=
sinθ
cosθ
​
=
1/4
15

​
/4
​
=
15

​

Substitute these values into the required expression:

cscθ+cosθ+cotθ=4+
4
15

​

​
+
15

​

=4+
4
15

​
+4
15

​

​

=4+
4
5
15

​

​
=
4
16+5
15

​

​