Question

in a right angled triangle ABC , right angled at B . if tanA = 1/√3 . then find the value of COSACOSC - SINASINC

Answer 1

Answer:

Explanation:

this is the required answer and solution

Answer 2

Answer:

Explanation:

Sorry Friends I Cannot Add Division Symbol in SOME Areas

In △ABC,

∠B=90o, tanA=3

​1​=ABBC​

Let BC =1x,AB=3

​x

AC2=AB2+BC2

AC2=(3

​x)2+(x)2=4x2

AC=2x

(i) sinAcosC+cosAsinC=21​×21​+23

​​×23

​​=4/1​+4/3​=1

(ii) cosAcosC−sinAsinC=23

​​×21​−21​×23

​​=4/3

​​−4/3

​​=0