In a right angled triangle ABC right angled at B given 15 cosA-8sín A=0 then
sin A + COS A
2 cos A-sinA
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1
Answer:
Given that
tanA=
1
3
AC=
(
3
)
2
+1
2
=
3+1
=
4
=2
now
(i)sinAcosC+cosAsinC
=
2
3
×
2
3
+
2
1
×
2
1
=
4
3
+
4
1
=
4
4
=1
and
(ii)cosAcosC−sinAsinC
=
2
1
×
2
3
−
2
1
×
2
3
=
4
3
−
4
3
=0
Step-by-step explanation:
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