Question

Evaluate, sec210° - cot2 80° + sin 15° cos 75° + COS 15° sin 75°
cos & sin(90 - 8) + sin 8 cos(90 - 8) *​

Answer 1

Step-by-step explanation:

Let's divide these into three parts :

♦ sec²(10) − cot²(80)

♦ sin(15) cos(75) + cos(15) sin(75)

♦ cosθ sin(90-θ) + sinθ cos(90-θ)

Simplifying Each :

♦ sec²(10) − cot²(80)

= 1/cos²(10) − cos²(80)/sin²(80)

= 1/sin²(80) − cos²(80)/sin²(80)

= (1−cos²(80))/sin²(80)

= sin²(80)/sin²(80)

= 1

♦ sin(15) cos(75) + cos(15) sin(75)

= sin(15+75)

= sin(90)

= 1

♦ cosθ sin(90-θ) + sinθ cos(90-θ)

= sin [ θ+(90−θ) ]

= sin(90)

= 1

Putting all them back together :

= 1 + 1 / 1

= 1 + 1

= 2