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11th Business Maths Monthly Test - 1 ( Trigonometry )

MABS Institution

11th Business Maths Monthly Test - 1 ( Trigonometry )-Aug 2020

Time : 2.30 Hours

Part - A

Multiple Choice Question - Answer All Question

The radian measure of 37^o30^' is

\(\frac{5\pi}{24}\)
\(\frac{3\pi}{24}\)
\(\frac{7\pi}{24}\)
\(\frac{9\pi}{24}\)

The value 4cos³40^o-3cos40^o is

\(\frac{\sqrt3}{2}\)
\(\frac{-1}{2}\)
\(\frac{1}{2}\)
\(\frac{1}{\sqrt2}\)

The degree measure of \(\frac{\pi}{8}\) is

20^o60^'
22^o30^'
20^o60^'
20^o30^'

\(\left(\frac{\cos x}{cosec x}\right)-\sqrt{1-\sin^2x}\sqrt{1-\cos^2x}\) is

cos²x-sin²x
sin²x-cos²x
1
0

\(\sin\left(\cos^{-1}\frac{3}{5}\right)\) is

\(\frac{3}{5}\)
\(\frac{5}{3}\)
\(\frac{4}{5}\)
\(\frac{5}{4}\)

If sin A+ cos A=1, then sin 2A is equal to

1
2
0
\(\frac{1}{2}\)

The value of \(\cos(-480^o)\) is

\(\sqrt3\)
\(-\frac{\sqrt3}{2}\)
\(\frac{1}{2}\)
\(\frac{-1}{2}\)

The value of sec A sin(270^o+A) is

-1
cos² A
sec² A
1

The value of cos²45^o-sin²45^o is

\(\frac{\sqrt3}{2}\)
\(\frac{1}{2}\)
0
\(\frac{1}{\sqrt{2}}\)

The value of sin 15^o cos 15^o is

1
\(\frac{1}{2}\)
\(\frac{\sqrt3}{2}\)
\(\frac{1}{4}\)

Part - B

Generic 2 - Answer All Question

Find the value of tan 15^o

Find the principal value of the following
\(\sec^{-1}(-\sqrt{2})\)

Find the degree measure corresponding to the following radian measure. \(\frac { \pi }{ 8 } \)

Find the values of the following trigonometric ratios. \(\cos { \left( { -210 }^{ o } \right) } \)
Find the quadrants in which the terminal sides of the following angles lie.
1325^o

Prove that \({ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 13 } \right) ={ tan }^{ -1 }\left( \frac { 2 }{ 9 } \right) \)

Show that \(\frac { sin2\theta }{ 1+cos2\theta } =tan\theta \)

Evaluate \(\cot\left(\frac{-15\pi}{4}\right)\)

Find the degree measure corresponding to the following radian measure. \(\frac { 11\pi }{ 18 } \)

Evaluate\(\cos\left[\frac{\pi}{3}-\cos^{-1}\left(\frac{1}{2}\right)\right]\)

Part - C

Generic 3 - Answer All Question

If tanA =\(\frac{1}{7}\) and tanB =\(\frac{1}{3}\), show that cos2A = sin4B

If tan(x + y) = 42 and x = tan–1(2), then find y

Convert the following into the product of trigonometric functions
sin7\(\theta\)-sin4\(\theta\)

Prove that \(\frac{\sin5x+\sin3x}{\cos5x+\cos3x}=\tan4x\)

Express \(\tan ^{ -1 } \left( \frac { \cos x-\sin x }{ \cos x+\sin x } \right) ,0<x<\pi \), in the simplest form.
Prove that tan 4A tan 3A tan A + tan 3A + tan A - tan 4A = 0

Find the values of the following sin (-105)°

If tan A = m tanB, prove that \(\frac { sin(A+B) }{ sin(A-B) } =\frac { m+1 }{ m-1 } \)

Prove that \(2\sin ^{ 2 }{ \frac { \pi }{ 6 } } +\ cosec ^{ 2 }{ \frac { 7\pi }{ 6 } } \cos ^{ 2 }{ \frac { \pi }{ 3 } } =\frac { 3 }{ 2 } \)

Express each of the following as the product of sine or cosine.
cos2ፀ-cosፀ

Part - D

Generic 5 - Answer All Question

If cosA =\(\frac{4}{5}\)and cosB =\(\frac{12}{13}\),\(\frac{3\pi}{3}\)\(\pi\), find the value of sin(A-B)

If \(\sin { \theta \frac { 3 }{ 5 } } \), \(\tan { \phi } =\frac { 1 }{ 2 } \)and \(\frac { \pi }{ 2 } <\theta <\pi<\varphi <\frac { 3\pi }{ 2 } \) , then find the value of \(8\tan { \theta } -\sqrt { 5 } \sec {\varphi } \)

If \(\sin { A } =\frac { 3 }{ 5 } \) where\(0<A<\frac { \pi }{ 2 } \) and \(\cos { B } =\frac { -12 }{ 13 } \) , \(\pi <A<\frac { 3\pi }{ 2 } \) , find the values of the following sin (A - B)

Prove that \({ cot }^{ -1 }\left[ \frac { \sqrt { 1+sinx } +\sqrt { 1-sinx } }{ \sqrt { 1+sinx } -\sqrt { 1-sinx } } \right] =\frac { x }{ 2 } \) where \(x\in \left( 0,\frac { \pi }{ 4 } \right) \)

Prove that cos²2x-cos²6x = sin 4x.sin 8x

If cosA+cosB=\(\frac { 1 }{ 2 } \) and sinA+sinB=\(\frac { 1 }{ 4 } \), prove that tan\(\left( \frac { A+B }{ 2 } \right) =\frac { 1 }{ 2 } \)

If \(\sin A=\frac35\) where \(0<A<\frac { \pi }{ 2 } \) and \(cosB=\frac { -12 }{ 13 } ,\pi <A<\frac { 3\pi }{ 2 } \) find the value of
\(\tan(A-B)\)
Find the value of tan \(\left( {{\pi}\over{8}}\right).\)

Prove that \(\frac{\cos4x+\cos3x+\cos2x}{\sin4x+\sin3x+\sin2x}=\cot3x\)

Prove that \(\frac { 4tan\ x(1-{ tan }^{ 2 }x) }{ 1-6{ tan }^{ 2 } x+{ tan }^{ 4 } x } =tanx\)

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