maths
posted by sakthi .
if tan theta= 13/12 where theta is in 2nd quadrant find remaing trigonometric ratios
Respond to this Question
Similar Questions

Maths complex numbers
Find tan(3 theta) in terms of tan theta Use the formula tan (a + b) = (tan a + tan b)/[1  tan a tan b) in two steps. First, let a = b = theta and get a formula for tan (2 theta). tan (2 theta) = 2 tan theta/[(1  tan theta)^2] Then … 
Trigonometric Functions
Determine the Value of all six trigonometric functions of theta when given? 
simple trig
If sin(theta)=[sqrt(70)]/7 and theta is in Quadrant two, find the exact numerical value of tan theta without using a calculator. I got tan(theta)=[sqrt(294)]/42 Is that right? 
easy trig
If sin(theta)=[sqrt(70)]/7 and theta is in Quadrant two, find the exact numerical value of tan theta without using a calculator. I got tan(theta)=[sqrt(294)]/42 Is that right? 
Trigonometry
I don't understand how I'm supposed set the problem up or what theta is... Use the given function value(s), and trigonometric identities (including the cofunction identities), to find the indicated trigonometric functions. sec theta … 
Math
1. Let (7, 4) be a point on the terminal side of (theta). Find the exact values of sin(theta), csc(theta), and cot(theta). 2. Let (theta) be an angle in quadrant IV such that sin(theta)=2/5. Find the exact values of sec(theta) and … 
trig
If sin theta is equal to 5/13 and theta is an angle in quadrant II find the value of cos theta, sec theta, tan theta, csc theta, cot theta. 
math
Another: theta is a second quadrant angle in standard position and tan theta =  (sqrt5)/(2) . Find the exact value of cos theta. I assume I use the equation 1 / cos(theta)^2 = 1 + tan (theta)^2 However how does this work out algebraically? 
Calculus
If cos of theta= 7/12 and theta is in quadrant 2 then find a. tan(theta) cot (theta)= b. csc(theta) tan(theta)= c. sin^2(theta)+ cos^2(theta)= 
Math: Trigonometry
Quadrant III, csc(theta)= 1.25, find tan(theta) Quadrant IV, sec(theta)= 4/3, find cot(theta) Please provide the work for both. Thank you!