Question

can someone answer it fast ​

Answer 1

Answer:

Cos58°cos32°-cos58°sin58°

(Cancelling cos58 & -cos58)we get

Cos32° - sin58°

[Sin58°= cos(90-58)]; 90-58=32

Cos32°-cos32°=0

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

$\leadsto {\tt {cos 58^{\circ}cos 32^{\circ}-cos 58^{\circ}sin 58^{\circ}}}$

By the formula,

$\leadsto\huge {\boxed {\tt {cos \theta = sin (90^{\circ}-\theta)}}}$

$\leadsto {\tt {cos 58^{\circ}sin (90-32^{\circ})-cos 58^{\circ}sin 58^{\circ}}}$

$\leadsto {\tt {cos 58^{\circ}sin 58^{\circ}-cos 58^{\circ}sin 58^{\circ}}}$

$\leadsto \huge {\boxed {\tt {0}}}$

$\rule{200}2$

Related Formulas :-

• ${\tt {cos \theta = sin (90^{\circ}-\theta)}}$
• ${\tt {sin \theta = cos (90^{\circ}-\theta)}}$
• ${\tt {tan \theta = cot (90^{\circ}-\theta)}}$
• ${\tt {cot \theta = tan (90^{\circ}-\theta)}}$
• ${\tt {cosec \theta = sec (90^{\circ}-\theta)}}$
• ${\tt {sec \theta = cosec (90^{\circ}-\theta)}}$

$\rule {200}2$