Simplify \(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\)
tan x
tanx secx
\(\sec^2 x\)
cosec2x
Correct answer is C
\(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\)
= \(\cos^{2} x \sec^{2} x + \cos^{2} x \sec^{2} x \tan^{2} x\)
= \(1 + \tan^{2} x\)
= \(1 + \frac{\sin^{2} x}{\cos^{2} x}\)
= \(\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}\)
= \(\frac{1}{\cos^{2} x} = \sec^{2} x\)
60 (tan 62o - tan 64o)
60 (cot 64o - cot 62o)
60 (cot 62o - cot 64o)
60 (tan 64o - tan 62o)
Correct answer is D
\(\frac{BC}{60}\) = \(\frac{tan 62}{1}\)
BC = 60 tan 62
\(\frac{AC}{60}\) = \(\frac{tan 62}{1}\)
AC = 60 tan 64
AB = AC - BC
= 60(tan 64o - tan 62o)
If the exterior angles of a pentagon are x°, (x + 5)°, (x + 10)°, (x + 15)° and (x + 20)°, find x
118o
72o
62o
36o
Correct answer is C
The sum of the exterior angles of a regular polygon = 360°.
\(\therefore x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360°\)
\(5x + 50 = 360° \implies 5x = 360° - 50° = 310°\)
\(x = \frac{310°}{5} = 62°\)
8\(\sqrt{3}\)
\(\frac{16}{\sqrt{3}}\)
\(\sqrt{3}\)
\(\frac{10}{\sqrt{3}}\)
Correct answer is A
\(\frac{x}{4}\) = \(\frac{\tan 60}{1}\)
x = 4 tan60
= 4\(\sqrt{3}\)
BD = 2x
= 8\(\sqrt{3}\)
The area of a square is 144 sq cm. Find the length of its diagonal
11\(\sqrt{3cm}\)
12cm
12\(\sqrt{2cm}\)
13cm
Correct answer is C
BD = \(\sqrt{x^2 + x^2}\)
= \(\sqrt{12^2 + 12^2}\)
= \(\sqrt{144 + 144}\)
= 2(144)
= 12\(\sqrt{2cm}\)