welcome back mathematicians in this video we are going to discuss five of the fundamental trigonometric identities and an identity is an equation that is true for all values of the variable in this instance theta can be replaced with any angle as long as the trigonometric function is defined for that angle and that equation would be true so let's start with the three reciprocal identities cosecant of angle theta is equal to one over sine of angle theta secant of angle theta is equal to 1 over cosine of angle theta and cotangent of angle theta is equal to 1 over tangent of angle theta our two quotient identities would be tangent of angle theta is equal to sine of angle theta over cosine of angle theta and cotangent of angle theta is equal to cosine of angle theta over sine of angle theta we will now use those identities to solve this problem says given sine of angle theta is equal to square root of 10 over 10 and cosine of angle theta is equal to 3 times the square root of 10 over 10 find the value of each of the four remaining trigonometric functions for angle theta so let's start with cosecant so we know cosecant of angle theta based on our reciprocal identity is equal to 1 over sine of angle theta because we know sine of angle theta is equal to square root of 10 over 10 we can then replace sine of angle theta with the ratio of square root of 10 over 10. we now need to simplify this and 1 over square root of 10 over 10 is equal to 10 over the square root of 10. we should always try to rationalize the denominator and to do that we're going to multiply both the numerator and denominator by square root of 10. this will then give us 10 times the square root of 10 over 10 which then can be reduced or simplified to be just square root of 10. we can now find secant of angle theta using really the same process so secant of angle theta is equal to one over cosine of angle theta we're going to replace cosine of angle theta with three times the square root of ten over ten we know that 1 over 3 times the square root of 10 over 10 is equal to 10 over 3 times the square root of 10. we should now rationalize the denominator by multiplying square root of 10 and square root of 10 to both the numerator and denominator this then gives us 10 times the square root of 10 over 3 times 10. we should reduce this fraction and in reducing it we get square root of 10 over 3. let's now find tangent of angle theta which we know based on the quotient identity is equal to sine of angle theta over cosine of angle theta we can now replace sine and cosine with their respective ratios so sine is equal to square root of 10 over 10 and cosine is equal to 3 times square root of 10 over 10. we're now going to simplify this ratio by taking the numerator and multiplying it by the reciprocal of the denominator which would be 10 over 3 times square root of 10. let's now reduce by common factors so we have a common factor of 10 in the numerator and denominator and a common factor of square root of 10 in the numerator and denominator this leaves us after we reduce with one third let's now find the cotangent of angle theta and we really have a choice in this instance we can actually do the quotient identity which means we would simplify using cosine of theta over sine of theta and it would be a pretty similar process to what we completed with the tangent of angle theta or we can use the reciprocal identity and say cotangent of angle theta is equal to one over tangent of angle theta well we know tangent of angle theta from the work we just completed is really one-third and one over one-third is three so now we've successfully found the ratios for the remaining four trigonometric functions using the fundamental identities good luck guys