If tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\) find sec\(\theta\)
\(\frac{m^2 + n^2}{(m^2 - n^2)}\)
\(\frac{m^2 + n^2}{2mn}\)
\(\frac{mn}{2(m^2 + n^2)}\)
\(\frac{m^2n^2}{2(m^2 - n^2)}\)
Correct answer is B
Tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\)
\(\frac{\text{Opp}}{\text{Adj}}\) by pathagoras theorem
= Hyp2 = Opp2 + Adj2
Hyp2 = (m2 - n2) + (2mn)2
Hyp2 = m4 - 2m2n4 - 4m2 - n2
Hyp2 = m4 + 2m2 + n2n
Hyp2 = (m2 - n2)2
Hyp2 = \(\frac{m^2 + n^2}{2mn}\)
The sine, cosine and tangent of 210o are respectively
\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{2}\)
\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{3}\)
\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{2}\)
\(\frac{-1}{2}\), \(\frac{\sqrt{-3}}{2}\), \(\frac{\sqrt{3}}{3}\)
Correct answer is D
210o = 180o - 210o = - 30o
From ratio of sides, sin -30o = -\(\frac{1}{2}\)
Cos 210o = 180o - 210o = -30o
= cos -30o = \(\frac{-3}{2}\)
But tan 30o = \(\frac{1}{\sqrt{3}}\), rationalizing this
= \(\frac{1}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{3}}{3}\)
∴ = \(\frac{-1}{2}\), \(\frac{\sqrt{-3}}{2}\), \(\frac{\sqrt{3}}{3}\)
35o
37o
49o
59o
Correct answer is D
In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59°
PRQ = QNP = 59°(angles in the same segment of a circle are equal)
210o
150o
105o
50o
Correct answer is D
Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the Nonagon = 9
Where 3 are equal and 6 other angles = 1110o
(2 x 9 - 4)90o = (18 - 4)90o
14 x 90o = 1260o
9 angles = 1260°; 6 angles = 110o
Remaining 3 angles = 1260o - 1110o = 150o
Size of one of the 3 angles \(\frac{150}{3}\) = 50o
Find the eleventh term of the progression 4, 8, 16.....
213
212
211
210
Correct answer is B
a = 4, r = \(\frac{4}{2}\)
\(\frac{8}{4}\) = 2
n = 11
Tn = arn - 1
T11 = 4(2)11 - 1
4(2)10 = 212
since 4 = 22
= 212