2010年8月13日 · I have trouble doing this problem. The problem: Let tan β= 6 Find tan(90°-β). Please and thank you.

2013年11月12日 · Yes, tan(x+ 90)= \frac{sin(x+ 90)}{cos(x+ 90)}= \frac{cos(x)}{-sin(x)}= -cot(x). (I am assuming that your "x" and "90" are in degrees since that is the only way this ...

sin -90 = - sin90= -1 cos-90=cos90=0 tan -90= -tan90= -oo cot-90= -cot90=0 csc-90=-csc90= -1/sin90= -1 sec-90= sec90= 1/cos90=oo

2016年4月10日 · The tangent of 90 degrees is not defined, and you can't divide by cot 90° which is zero.

2018年3月10日 · {: (sin(90^circ)=1,color(white)("xx"),sin(180^circ)=0), (cos(90^circ)=0,,cos(180^circ)=-1), (tan(90^circ)" is undefined",,tan(180^circ)=0) :} For the unit circle we have the situations pictured below; 90^circ and 180^circ are limits as the hypotenuse approaches the (positive) vertical axis and the (negative) horizontal axis respectively.

2018年4月29日 · Given that cosx=p Where x us an acute angle in degrees find in term of p i) Sin x ii)tanx iii)tan(90-x)?

2018年3月26日 · Please see below. sin(-90^0)==-sin90^0=-1 cos(-90^0)==cos90^0=0 tan(-90^0)==-tan90^0= undefined csc(-90^0)==-csc90^0=-1 sec(-90^0)==sec90^0= undefined cot(-90^0 ...

2016年6月10日 · How do you find all the solutions for #2 \sin^2 \frac{x}{4}-3 \cos \frac{x}{4} = 0# over the

2015年4月15日 · Probably not the answer you were hoping for but If tan(x) = 90 then arctan(tan(x)) = arctan(90) x = arctan(90) (I am not aware of any simpler version than this.)

2018年6月27日 · =-cosx The sine function is periodic with period 360^@, hence sin(360^@-x) =sin(-x)=-sinx About the tangent function; let y=90^@ - x => x+y=90^@. x and y are complementary angles. We usually meet them in right triangles, where trigonometric functions are usually more familiar. As such, let x and y be the acute angles of a right triangle with sides cc …