Question

OS Q)
0.
29. In a AABC, B =90 and tan A = 1
that
(i) sin A.cos C+cos A sin C=1
(ii) cos A.cos C-sin A sin C = 0​

Answer 1

Step-by-step explanation:

a AABC, B =90 and tan A = 1

here A=45

then in a triangle A+B+C=180

C=45

i) sin A.cos C+cos A sin C

sin(A+C)=sin(45+45)=sin90=1

(ii) cos A.cos C-sin A sin C

cos(A+C)=cos90=0

Answer 2

Step-by-step explanation:

In ∆ABC ,

B = 90° , tanA = 1/1 = लम्ब/आधार

कर्ण = √2

SinA = 1/√2 , CosA = 1/√2

SinC = 1/√2 , CosC = 1/√2

(i) 1/√2 * 1/√2 + 1/√2 * 1/√2

= 1/2 + 1/2 = 2/2 = 1

(ii) 1/√2 * 1/√2 - 1/√2 * 1/√2

= 1/2 - 1/2 = 0