tan x cosec x
-cot x cosec x
tan x sec x
-cot x sec x
Correct answer is B
\(\csc x = \frac{1}{\sin x}\)
Using the quotient rule,
\(\frac{\mathrm d y}{\mathrm d x} = \frac{vdu - udv}{v^{2}}\)
= \(\frac{\sin x (0) - 1 (\cos x)}{(\sin x)^{2}}\)
= \(\frac{- \cos x}{\sin^{2} x}\)
= \((\frac{- \cos x}{\sin x}) (\frac{1}{\sin x})\)
= \(- \cot x \csc x\)
60m
30 \(\sqrt{3}\)m
20 \(\sqrt{3}\)m
10 \(\sqrt{3}\)m
Correct answer is B
h = 30 tan 60°
= 30\(\sqrt{3}\)
100o
80 o
50 o
40 o
Correct answer is B
< PYU = < YPQ = 40°(Alternative) = PQ\(\parallel\)UV
< XPQ = < YPQ = 40°, Since QX = QY
therefore < XPY = 80°
x + 2y - 1= -0
2x + y - 3 = 0
x - 2y - 3 = 0
2x + y - 1 = 0
Correct answer is D
Line PQ : x - 2y + 4 = 0
2y = x + 4 \(\implies y = \frac{x}{2} + 2 \)
Slope = \(\frac{1}{2}\)
Slope of the perpendicular line QR: \(\frac{-1}{\frac{1}{2}} = -2\)
Line QR: \(y = mx + b\)
\(y = -2x + b\)
Point of intersection: (1, -1)
\(-1 = -2(1) + b \implies b = -1 + 2 = 1\)
\(y = -2x + 1 \implies y + 2x - 1 = 0\)
\(QR: 2x + y - 1 = 0\)
\(\frac{1100}{3}\)km
\(\frac{2200}{3}\)km
\(\frac{22000}{3}\)km
\(\frac{1100}{3}\)km
Correct answer is C
Angular difference = 45° + 15° = 60°.
\(D = \frac{60°}{360°} \times 2 \times \frac{22}{7} \times 7000\)
= \(\frac{22000}{3} km\)