how do i prove this identity sin^2C (1+cot^2C) + cos^2C (1+tan^2C) = 2 sorry i suck at this?

Question

i have already asked an identities question today....but i just cannot do them! please help. on and the ^2 means squared!

Answer 1

One important rule is that cos^2+sin^2=1

First of all, do this:
sin^2c(1+(cos^2c/sin^2c))
+cos^2c(1+(sin^2c/cos^2c))=
sin^2c+cos^2c+cos^2c+sin^2c=
cos^2c+sin^2c+cos^2c+sin^2c=
1+1=2
2=2

I hope this helps!

Answer 2

There is only one identity you need to know to answer this question, that sin^2 C + cos^2 C = 1

You can further transform this equation to help you more effectively.
1. Divide the entire eqn by sin^2 C:
1 + cot^2 C = cosec^2 C = 1/(sin^2 C)
2. Divide the entire eqn by cos^2 C:
1 + tan^2 C = sec^2 C = 1/(cos^2 C)

sub this into your question, and you immediately get sin^2C*1/(sin^2 C)+ cos^2C*1/(cos^2 C) = 1 + 1 = 2 (PROVEN)

Hope this helps=)

Answer 3

as you know :
1+cot^2C= 1/sin^2C
and
1+tan^2C= 1/cos^2C
so
sin^2C (1+cot^2 C) + cos^2C (1+tan^2C)=
sin^2C (1/sin^2C) + cos^2 C (1/cos^2 C)=
1+1=2

Answer 4

Just remember, whenever possible, simplify first. You can do this by converting all trig functions to sine and cosine functions, then apply the rules of mathematics and trigonometry.

For example, the cotangent function can be re-written as cos(x)/sin(x), and the tangent function can be re-written as sin(x)/cos(x). In your particular problem, after re-writing these functions, just apply the distributive property and multiply. Then collect like terms. Then factor the result. One of those factors should be sin^2(x) + cos^2(x), whose sum just happens to be 1 for any angle (x). This should give you enough information to complete your assignment on your own.

Answer 5

LHS= sin^2c[1 + cos^2c/sin^2c] + cos^2c[1 + sin^2c/cos^2c]
= sin^2c[1/sin^2c] + cos^2c[1/cos^2c) ;(took lcm)
= 1+ 1
=2=RHS