Let 0° < t < 90°. Then which of the followings is true?

Correct option is C

Given:

0° < t < 90°

Formula Used:

​Solution:

We can see from the table that $$\sin 0° =0 \ and\ \sin 90° =1$$​​

So, value of sine increases from 0 to 1 when angle increases from 0° to 90°

And for cosine:

The value of $$\cos 0° =1 \ and\ \cos 90° =0$$​​

So, the value of cosine decreases from 1 to 0 when angle increases from 0° to 90°

From table:

​$$\sin 45° = \frac{1}{\sqrt{2}}$$​​

​$$\cos 45° = \frac{1}{\sqrt{2}}$$​​

At 0° sine is less than cosine

At 90° cosine is less than sine

And at t = 45° both sine and cosine have same value

So, we can say that for:

0° < t < 45°

​$$\cos (t) >\sin(t)$$​​

45° < t < 90°

​$$\cos (t) < \sin(t)$$​​

So, option (3) sin(t) < cos (t) when t < 45° is correct