Question

Prove that : sin80°-cos70°=cos50°

Answer 1

SOLUTION

=) sin80°- cos70°= cos50°

=) sin 80°= cos50°+ cos70°

Take R.H.S:

=)cos50° +cos70°

$= > (</strong><strong>cosA</strong><strong> + </strong><strong>cosB</strong><strong> = 2cos \frac{</strong><strong>A</strong><strong>+ </strong><strong>B</strong><strong>}{2} cos \frac{</strong><strong>A</strong><strong> - </strong><strong>B</strong><strong>}{2} ) \\ \\ = > 2cos \frac{50 \degree + 70 \degree}{2}cos \frac{50 \degree - 70 \degree}{2} \\ \\ = > 2cos \frac{120 \degree}{2} cos \frac{ - 20 \degree}{2}$

=) 2cos 60° cos(-10°)

=){cos(-A)= cosA}

=) 2× 1/2× cos 10°

=) cos10°

=) cos(90°-80°)

=) {cos(90°-A)= sin A}

=) sin 80° L.H.S

Hence proved

Answer 2

Step-by-step explanation:

Consider LHS:

sin⁡80∘−cos⁡70∘

=sin⁡80∘−cos⁡(90∘−20∘)

=sin⁡80∘−sin⁡20∘

=2sin⁡(80∘−20∘2)cos⁡(80∘+20∘2){∵sin⁡A−sin⁡B=2sin⁡(A−B2)cos⁡(A+B2)}

=2sin⁡30∘cos⁡50∘

=2×12cos⁡50∘

=cos⁡50∘

= RHS

Hence, LHS = RHS.