Question

In triangle ABC if LB=90 and cot A=1, then the value of cos C Sin A-Cos A Sin c is

Answer 1

Answer:

Given in △ ABC, ∠B=90o.

Given in △ ABC, ∠B=90o.This given A+C=180o−90o=90o......(1).

Given in △ ABC, ∠B=90o.This given A+C=180o−90o=90o......(1).Now,

Given in △ ABC, ∠B=90o.This given A+C=180o−90o=90o......(1).Now,sinA.cosC+cosA.sinC

Given in △ ABC, ∠B=90o.This given A+C=180o−90o=90o......(1).Now,sinA.cosC+cosA.sinC=sin(A+C)

Given in △ ABC, ∠B=90o.This given A+C=180o−90o=90o......(1).Now,sinA.cosC+cosA.sinC=sin(A+C)=sin(90o)

Given in △ ABC, ∠B=90o.This given A+C=180o−90o=90o......(1).Now,sinA.cosC+cosA.sinC=sin(A+C)=sin(90o)=1.

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Answer 2

Answer:

A+B+C=180

in triangle

A+C=180-90

A+C=90

cosCsinA-CosAsinC=cos(A+B)

cos90=0