Question

cosA=3/5 எ‌னி‌ல் (sinA-cosA)/2tanA இ‌ன் ம‌தி‌ப்பை‌க் கா‌ண்க

Answer 1

Answer:

3/40

Step-by-step explanation:

H=5,B=3

By pythagoras theorme we get,

P=4 , Then ,

(sinA-cosA)/2tanA

(P/H - B/H)/2P/B

(4/5 - 3/5)/2×4/3

( 4-3)/5 × 3/8

1/5×3/8

3/40

Answer 2

விளக்கம்:

$\begin{aligned}&B C=\sqrt{A C^{2}-A B^{2}}\\&=\sqrt{5^{2}-3^{2}}\\&=\sqrt{25-9}\\&=\sqrt{16\\}\\ & =4\end{aligned}$

$\sin A=\frac{\text {opp side}}{\text {hypo}}$

$=\frac{4}{5}$

$\cos A=\frac{adjside}{hypo }$

$\cos A=\frac{3}{5 }$

$\tan A=\frac{\text {opp.side}}{\text {adj} \text {side}}=\frac{4}{3}$

$\frac{\sin A-\cos A}{2 \tan A}$ ன் மதிப்பு

$\frac{\frac{4}{5}-\frac{3}{5}}{2\left(\frac{4}{3}\right)}$

$\frac{\frac{4-3}{5}}{\frac{8}{3}} = \frac{\frac{1}{5}}{\frac{8}{3}}$        

$=>\frac{1}{5} \times \frac{3}{8}$

$=\frac{3}{40}$

$\frac{\sin A-\cos A}{2 \tan A}=\frac{3}{40}$