Question

evaluate sec²10-cot²80 +sin15cos75+cos15sin75/cosx sin (90-x)+sinxcos(90-x)​

Answer 1

Answer:

Didn't understand ur ques

Answer 2

Answer:

Step-by-step explanation:Hi there!

We have,

sec²(10) − cot²(80) + sin(15) cos(75) + cos(15) sin(75)/ cosθ sin(90-θ) + sinθ cos(90-θ) 

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

[ Thank you! for asking the question. ]

Hope it helps!