Question

26. If in ∆ABC,
<B = 90° and tan C
1/2
then find the value of sin A ×cos C - Cos A ×sin C.​

Answer 1

This is your required answer .Hope It will help you

Answer 2

Answer:

Step-by-step explanation:

given , ∆ABC, ∠ B = 90° and tan C = $\frac{1}{2}$

make a right angle triangle tan c = $\frac{1}{2} = \frac{P}{B}$

let AB (P) = 1K , BC (B) = 2K and AB = h

[ where P = perpendicular and B = base ]

by using pythagoras theorem

(AC)² = (BC)² + (AB)²

(h)² = (2K)² + (1K)²

      = 4K² + K²

      = 5 K²

      = $\sqrt{5} K$

sin C = $\frac{1K}{\sqrt{5} K}$ = $\frac{1}{\sqrt{5} }$

cos C = $\frac{2K}{\sqrt{5} K}$ = $\frac{2}{\sqrt{5} }$

sin A  = $\frac{2K}{\sqrt{5}K }$ = $\frac{2}{\sqrt{5} }$

cos A = $\frac{1K}{\sqrt{5} K}$ = $\frac{1}{\sqrt{5} }$

sin A × cos C - cos A × sin C

$=\frac{2}{\sqrt{5} }$ × $\frac{2}{\sqrt{5} }$ $- \frac{1}{\sqrt{5} }$ × $\frac{1}{\sqrt{5} }$

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

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

= $\frac{3}{5}$