Question

A `DeltaABC` is right angled at `B`. Find the value of?

`(SecA.SinC - tanA.tanC)/sinB`

Answer 1

Zero.

Explanation:

Use the trigonometric relations for complementary angles in the numerator, as the acute angles A and C in the triangle are complementary:

`\sec A \sin C=(\sin C)/(\cos A)=(\sin C)/(\sin C)=1`

`\tan A \tan C=(1/(\tan C))(\tan C)=1`

Then

`\sec A \sin C-\tan A \tan C=1-1=0`

The denominator is `\sin B=1` as angle B is a right angle. So, the given fraction is `0/1=0`.