Question

If tan 0 =12/13 then
sin0+ cos0/cos0-sin0=

(b)23
(a)25
(c)22
(d) None of these​

Answer 1

Answer:

25

Step-by-step explanation:

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Tan(∅) = perpendicular/base = p/b

Let,

p=12k and b=13k

by using pythagoras theorem,

h² = p² + b²

h² = (12k)² + (13k)²

h² = 144k² + 169k²

h= √313 k

Now,

Sin(∅) = p/h = 12k/√313 k =12/√313

Cos(∅) = b/h = 13k/√313 k = 13/√313

Now substituting these values in question, we get

Sin∅+cos∅/Cos∅ - sin∅

=(12/√313 + 13/√313) / (13/√313 - 12/√313)

=(25/√313)/(1/√313)

=25