Question

b) sin 62º – cos 28°trigonommetry​

Answer 1

$\huge{\mathcal{\blue{\underline{\underline{\pink{HeYa!!!}}}}}}$

${\large{\purple{\mathbb{ANSWER}}}}$

☆ To find:-

• The value of Sin62°-Cos28°.

☆ Id(s) Used:-

• SinA = Cos (90-A)

• CosA = Sin (90-A)

☆Now,

=》sin 62º – cos 28°

= Cos (90°-62°) - Cos 28°

= Cos 28° - Cos 28°

${\huge{\blue{\boxed{=\:\:0}}}}$

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Answer 2

$\huge{\underline{\underline{Solution:-}}}$

sin62° - cos28°

= sin(90° - 28°) - cos28°

= cos28° - cos28°

= 0