In Exercises 17–20, θ is an acute angle and sin θ and cos θ are g...

Question

In Exercises 17–20, θ is an acute angle and sin θ and cos θ are given. Use identities to find tan θ, csc θ, sec θ, and cot θ. Where necessary, rationalize denominators.
__
sin θ = 6, cos θ = √13
7 7

Accepted Answer

Hello everybody. I hope you're doing. All right. Today, today we're going to begin this question that states an acute angle, alpha and trigonometric functions sign of alpha and co sign of alpha are given. And we are asked to find the remaining trigonometric functions. Tangents of alpha, cotangent of alpha sins alpha and cosin alpha by using identities. Now they tell us that sign of alpha is six divided by 11 and co sign of alpha is equal to the square root of 85 divided by 11. So using the identities, we'll first start with tangent. So we have that tangent of alpha is equal to sine of alpha divided by cosine of alpha. So we know that tangent of an angle is equal to sign of that angle divided by a coin of the angle. So if we apply it to our question, we'll have tangents of alpha is equal to sign of alpha. And the numerators will have open parentheses, six divided by 11 closed parentheses, divided by open parentheses, the square roots of 85 divided by 11 closed parentheses. Now this will simplify, the elevens will cancel out and this will simplify to have six divided by the square of 85. But we don't want to keep a square root in our denominator. So we'll multiply both the numerator and the denominator by the screw roots of 85. And we have that the tangent of alpha is equal to six, multiplied by the screwed of 85 divided by 85. And that will be our first answer. And now we can move on to co sequins. So we have code sequences of alpha is equal to one divided by sign of alpha. So cos is just one divided by the sign. So we'll have this will be equal to one divided by the sign of alpha, which is six divided by 11. So all we do is flip the sign that that was given to us and we'll have that cos second alpha is 11 divided by six and that will be our second answer. And then we can move on to second alpha. Now sees is similar to co can where cos was one divided by sign. Now we have one divided by cosine of alpha. So we applying the same concept, we'll have one divided by the square root of 85 divided by 11. So we just flip the cosine that was given to us and we'll have divided by the square root of 85. But once again, we don't want a square root in the denominator. So we'll multiply both numerator and denominator by the squared of 85 to leave us with 11 multiplying the square root of 85 divided by 85. And that will be our third and our answer for seconds. And finally, we'll look at co tangent of alpha and similar to C and C can coag will be one divided by the tangents mobile phone and we'll have one divided by six divided by the screwed of which means we just take that fraction and flip it and we'll have the square root of 85 divided by six. Now, when it came to code tangent, I was able to use the tangent that we had obtained of six divided by the square of 85 instead of using the final tangent. Because if we flip that tangent with the screw at the denominator, if we flip it, flip it, we'll have a screwed in the numerator, which is fine. So that's why I used that one for the code tangents and just like that by using the relationships between the trigonometric functions we've been able to solve for other functions. So I hope that was helpful. And until next time.

In Exercises 17–20, θ is an acute angle and sin θ and cos θ are given. Use identities to find tan θ, csc θ, sec θ, and cot θ. Where necessary, rationalize denominators. __ sin θ = 6, cos θ = √13 7 7

Key Concepts

Trigonometric Ratios

Reciprocal Identities

Rationalizing Denominators

Watch next