6a. 1 - 2sin(x)² - 2cos(x)² = 1 - 2(sin(x)² +cos(x)²) = 1 - 2·1 = -1
6c. tan(x) + sin(x)/cos(x) = tan(x) + tan(x) = 2tan(x)
6e. 3sin(x) + tan(x)cos(x) = 3sin(x) + (sin(x)/cos(x))cos(x) = 3sin(x) +sin(x) = 4sin(x)
6g. 1 - cos(x)²tan(x)² = 1 - cos(x)²·(sin(x)²)/cos(x)²) = 1 -sin(x)² = cos(x)²
In order to form triangle PQT and quadrilateral TQRS, point T must lie on line PS which is 16 cm. long.
If the ratio of PT to TS is 5:3 and the total length of PS is 16, then PT must be 10 and TS must be 6 (10 + 6 =16) and 10:6 is the same ratio as 5:3. Another way to think about it is 5/3 = 10/6.
Now you have all the lengths that you need to find the areas of the quadrilateral and the triangle.
Make sure you draw a diagram of it!!
Yes the first one is the correct answer in this form of a graph
