If cos\(\theta\) = \(\frac{a}{b}\), find 1 + tan2\(\theta\)
\(\frac{b^2}{a^2}\)
\(\frac{a^2}{b^2}\)
\(\frac{a^2 + b^2}{b^2 - a^2}\)
\(\frac{2a^2 + b^2}{a^2 + b^2}\)
Correct answer is A
cos\(\theta\) = \(\frac{a}{b}\), Sin\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a}}\)
Tan\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a^2}}\), Tan 2 = \(\sqrt{\frac{b^2 - a^2}{a^2}}\)
1 + tan2\(\theta\) = 1 + \(\frac{b^2 - a^2}{a^2}\)
= \(\frac{a^2 + b^2 - a^2}{a^2}\)
= \(\frac{b^2}{a^2}\)
At what points does the straight line y = 2x + 1 intersect the curve y = 2x2 + 5x - 1?
(-2, -3) and(\(\frac{1}{2}\), 2)
(1, 0), (1, 3)
(4, 0) and (0,1)
(2, 0) and (0,1)
Correct answer is A
No explanation has been provided for this answer.
3k2
2k - k2
\(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)
k2
Correct answer is C
\(\frac{a}{c}\) = \(\frac{c}{d}\) = k
∴ \(\frac{a}{b}\) = bk
\(\frac{c}{d}\) = k
∴ c = dk
= \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\)
= \(\frac{3(bk)^2 - (bk)(dk) + dk^2}{3b^2 - bd + a^2}\)
= \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)
k = \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)
Find the values of x which satisfy the equation 16x - 5 x 4x + 4 = 0
1 and 4
-2 and 2
0 and 1
-1 and 0
Correct answer is C
16x - 5 x 4x + 4 = 0
(4x)2 - 5(4x) + 4 = 0
let 4x = y
y2 - 5y + 4 = 0
(y - 4)(y - 1) = 0
y = 4 or 1
4x = 4
x = 1
4x = 1
i.e. 4x = 4o, x = 0
∴ x = 1 or 0
Make U the subject of the formula S = \(\sqrt{\frac{6}{u} - \frac{w}{2}}\)
u = \(\frac{12}{2s^2}\)
u = \(\frac{12}{2s+ w}\)
u = \(\frac{12}{2s^2 + w}\)
u = \(\frac{12}{2s^2}\) + w
Correct answer is C
S = \(\sqrt{\frac{6}{u} - \frac{w}{2}}\)
S = \(\frac{12 - uw}{2u}\)
2us2 = 12 - uw
u(2s2 + w) = 12
u = \(\frac{12}{2s^2 + w}\)