Question

If in triangle ABC angle C=90°and tan A=3/4 prove sinAcosB+cosA sinB=1

Answer 1

Step-by-step explanation:

the second triangle is for sinB and cosB

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Answer 2

Answer:

$\sin A \cos C+\cos A \sin C=1$is proved.

Given Data:

According to the Question:

If in triangle ABC angle C=90° and tan A$=\frac{3}{4}$

The triangle is attached below:

To Find:

$\sin A \cos B+\cos A \sin B=1$

Step-by-step explanation:

Step 1:

$\{Tan} A=\frac{B C}{A B}=\frac{1}{\sqrt{3}}$

Thus $B C=1 A B=\sqrt{3}$

In triangle ABC,

Step 2:

$A B^{2}+B C^{2}=A C^{2}$

Step 3:

Thus AC=2 (BY PYTHAGORAS THEORM)

Step 4:

$\sin A \cos C+\cos A \sin C=\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$

Step 5:

Result:

$\sin A \cos C+\cos A \sin C=\frac{1}{4}+\frac{3}{4}$

$\sin A \cos C+\cos A \sin C=\frac{4}{4}$

$\sin A \cos C+\cos A \sin C=1$