Question

Verify the identity

Answer 1

\(\Large\tan\theta+\cot\theta=\frac{1}{\sin\theta\cos\theta}\)

Answer 2

@ganeshie8

Answer 3

\[\large tan\theta + cot\theta = \frac{1}{sin\theta cos\theta}\]

Change everything to sines and cosines

\[\large \frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = \frac{1}{sin\theta cos\theta}\]

Now we need a common denominator...what would that be?

Answer 4

\(\sin\theta+\cos\theta\)?

Answer 5

Not quite....if we multiply the 2 denominators we get \(\large sin\theta cos\theta\) that that would be our common denominator...so what we need to do is

\[\large \frac{sin\theta}{sin\theta} \times \frac{sin\theta}{cos\theta} + \frac{cos\theta}{cos\theta} \times \frac{cos\theta}{sin\theta} = \frac{1}{sin\theta cos\theta}\]

Simplify that down a bit

\[\large \frac{sin^2 \theta}{sin\theta cos\theta} + \frac{cos^2 \theta}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\]

Now we can combine those 2 over the common denominator

\[\large \frac{sin^2 \theta + cos^2 \theta}{sin\theta cos\theta} = \frac{1}{sin\theta cos \theta}\]

now what else do we have to do?

Answer 6

multiply both sides by sinthetacostheta?

Answer 7

Not nearly as difficult...what is the mother of all trigonometric identities that we should know?

\[\large sin^2 \theta + cos^2 \theta = 1\]

Answer 8

is that what you get when you multiply both sides by the denominators?

Answer 9

Nope...so what we had was

\[\large \frac{sin^2 \theta + cos^2 \theta}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\]

On the left one...we see we have \(\large sin^2 \theta + cos^2 \theta\)

which we now know equals 1...so now we have

\[\large \frac{1}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\]

and the identity is confirmed!

Answer 10

what?

Answer 11

Let me gather everything together...maybe it will help you see it

\[\large tan\theta + cot\theta = \frac{1}{sin\theta cos\theta}\]

Changing everything to sines and cosines

\[\large \frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = \frac{1}{sin\theta cos\theta}\]

Finding the common denominator...which is \(\large sin\theta cos\theta\) ...we need to multiply the first fraction by sin/sin and the second fraction by cos/cos to even things out

\[\large \frac{sin\theta}{sin\theta} \times \frac{sin\theta}{cos\theta} + \frac{cos\theta}{cos\theta} \times \frac{cos\theta}{sin\theta} = \frac{1}{sin\theta cos\theta}\]

Simplify those products

\[\large \frac{sin^2\theta}{sin\theta cos\theta} + \frac{cos^2 \theta}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\]

Now that the two fractions have a common denominator...we combine them over that

\[\large \frac{\color \red{sin^2\theta + cos^2 \theta}}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\]

Now....we remember the Trigonometric Identity that states \(\large \color\red{sin^2\theta + cos^2 \theta = 1}\) so we use that here...

\[\large \frac{1}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}\]

Which verifies the identity

Answer 12

ohhh okay