Question #4aae2 Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer P dilip_k Feb 12, 2017 #y=(Sin4x+cos4x)/(sin4x-cos4x)# #=>y/1=(sin4x+cos4x)/(sin4x-cos4x)# By componendo and dividendo #=>(y+1)/(y-1)=(2sin4x)/(2cos4x)# #=>(y+1)/(y-1)=tan4x# #=>(y+1)/(y-1)=(2tan2x)/(1-tan^2 2x# Again by componendo and dividendo #=>(2y)/2=(2tan2x+1-tan^2 2x)/(2tan2x-1+tan^2 2x)# #=>y=((4tanx)/(1-tan^2x)+1-(4tan^2x)/(1-tan^2x)^2)/((4tanx)/(1-tan^2x)-1+(4tan^2x)/(1-tan^2x)^2)# Answer link Related questions What does it mean to find the sign of a trigonometric function and how do you find it? What are the reciprocal identities of trigonometric functions? What are the quotient identities for a trigonometric functions? What are the cofunction identities and reflection properties for trigonometric functions? What is the pythagorean identity? If #sec theta = 4#, how do you use the reciprocal identity to find #cos theta#? How do you find the domain and range of sine, cosine, and tangent? What quadrant does #cot 325^@# lie in and what is the sign? How do you use use quotient identities to explain why the tangent and cotangent function have... How do you show that #1+tan^2 theta = sec ^2 theta#? See all questions in Relating Trigonometric Functions Impact of this question 1270 views around the world You can reuse this answer Creative Commons License