Solve cos(-60^o)-tan135^o-:tan315^o +cos660^o? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Shwetank Mauria Jul 3, 2016 cos(-60^o)-tan135^o-:tan315^o +cos660^o=0 Explanation: To solve cos(-60^o)-tan135^o-:tan315^o +cos660^o we make use of cos60^o=1/2 and tan45^o=1 and also the identities cos(-A)=cosA and tan(360^o-A)=tan(180^o-A)=-tanA. Hence cos(-60^o)-tan135^o-:tan315^o +cos660^o = cos60^o-tan(180^o-45^o)-:tan(360^o-45^o) +cos(720^o-60^o) = cos60^o-(-tan45^o)-:(-tan45^o) +cos(-60^o) = 1/2-(-1)-:(-1) +cos60^o = 1/2-1+1/2=0 Answer link Related questions How do you find the trigonometric functions of any angle? What is the reference angle? How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle? What is the reference angle for 140^\circ? How do you find the value of cot 300^@? What is the value of sin -45^@? How do you find the trigonometric functions of values that are greater than 360^@? How do you use the reference angles to find sin210cos330-tan 135? How do you know if sin 30 = sin 150? How do you show that (costheta)(sectheta) = 1 if theta=pi/4? See all questions in Trigonometric Functions of Any Angle Impact of this question 1921 views around the world You can reuse this answer Creative Commons License