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

Question

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

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Answer 1 · 2018-02-04T21:28:21

Zero.

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 $0/1=0$ .