Question

sin A = sin (90°- B)
since A, B are acute angles, A=900 - B
A+B=90°
°
ple-Il. Express sin 81° + tan 81° in terms of trigonometric ratios of angles between
don: We can
and tan 81° = tan(90° - 99) = cot 9°
Then, sin 81° + tan 81° = cos 9° + cot 9°
B+C
2
pl-12. IfA, B and C are interior angles of triangle ABC, then show that sin (
뜰
А
=COS-
2
; Given A, B and C are angles of triangle ABC then A+B+C = 180°​

Answer 1

Answer:

sin A = sin (90°- B)

since A, B are acute angles, A=900 - B

A+B=90°

°

ple-Il. Express sin 81° + tan 81° in terms of trigonometric ratios of angles between

don: We can

and tan 81° = tan(90° - 99) = cot 9°

Then, sin 81° + tan 81° = cos 9° + cot 9°

B+C

2

pl-12. IfA, B and C are interior angles of triangle ABC, then show that sin (

뜰

А

=COS-

2

; Given A, B and C are angles of triangle ABC then A+B+C = 180°

Step-by-step explanation:

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