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Maths 1 Marks Question

Introduction to Trigonometry · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 20 questions.

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Question 11 Mark
Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $(cosec A - sin A) (sec A - cos A) =$ $\frac { 1 } { \tan A + \cot A }$
$[$Hint: Simplify $LHS$ and $RHS$ separately$]$
Answer
$LHS$
$= ( \csc A - \sin A ) ( \sec A - \cos A )$
$= \left( \frac { 1 } { \sin A } - \sin A \right) \left( \frac { 1 } { \cos A } - \cos A \right) = \frac { 1 - \sin ^ { 2 } A } { \sin A } \frac { 1 - \cos ^ { 2 } A } { \cos A }$
$= \frac { \cos ^ { 2 } A } { \sin A } \frac { \sin ^ { 2 } A } { \cos A } , \ldots \ldots \ \because \sin ^ { 2 } A + \cos ^ { 2 } A = 1 = \frac { \frac { \sin A \cos A } { \sin A \cos A } } { \frac { \sin ^ { 2 } A } { \sin A \cos A } + \frac { \cos ^ { 2 } A } { \sin A \cos A } }$
........ Dividing the numerator and denominator by sin A cos A
$= \frac { 1 } { \tan A + \cot A }$
$= RHS$
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Question 21 Mark
Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $\sqrt { \frac { 1 + \sin A } { 1 - \sin A } }$ $= sec A + tan A$
Answer
$L.H.S.$ $\sqrt { \frac { 1 + \sin A } { 1 - \sin A } }$
$= \sqrt { \frac { 1 + \sin A } { 1 - \sin A } } \times \sqrt { \frac { 1 + \sin A } { 1 + \sin A } }$
$=\sqrt { \frac { ( 1 + \sin A ) ^ { 2 } } { 1 - \sin ^ { 2 } A } } \left[ \because ( a + b ) ( a - b ) = a ^ { 2 } - b ^ { 2 } \right]$
$=\sqrt { \frac { ( 1 + \sin A ) ^ { 2 } } { \cos ^ { 2 } A } } \left[ \because 1 - \sin ^ { 2 } \theta = \cos ^ { 2 } \theta \right]$
$=\frac { 1 + \sin A } { \cos A } = \frac { 1 } { \cos A } + \frac { \sin A } { \cos A } = \sec A + \tan A = R \cdot H . S .$
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Question 31 Mark
Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $\frac { 1 + \sec A } { \sec A } = \frac { \sin ^ { 2 } A } { 1 - \cos A }$
$[$Hint: Simplify $LHS$ and $RHS$ separately$]$
Answer
$\mathrm { LHS } = \frac { 1 + \sec A } { \sec A } = \frac { 1 + \frac { 1 } { \cos A } } { \frac { 1 } { \cos A } }$
$= \frac { \frac { \cos A + 1 } { \cos A } } { \frac { 1 } { \cos A } } = \cos A + 1 = 1 + \cos A$
$= \frac { ( 1 + \cos A ) ( 1 - \cos A ) } { 1 - \cos A } = \frac { 1 - \cos ^ { 2 } A } { 1 - \cos A }$
$= \frac { \sin ^ { 2 } A } { 1 - \cos A } \cdot \because \sin ^ { 2 } A + \cos ^ { 2 } A = 1$
$= RHS$
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Question 41 Mark
Prove the given identity, where the angles involved are acute angles for which the expressions are defined.
$\left( \frac { 1 + \tan ^ { 2 } A } { 1 + \cot ^ { 2 } A } \right) = \left( \frac { 1 - \tan A } { 1 - \cot A } \right) ^ { 2 } = tan^2 A$
Answer
$= \frac { 1 + \tan ^ { 2 } A } { 1 + \cot ^ { 2 } A } = \frac { 1 + \tan ^ { 2 } A } { 1 + \frac { 1 } { \tan ^ { 2 } A } } \cdot \because \cot A = \frac { 1 } { \tan A }$
$= \frac { 1 + \tan ^ { 2 } A } { \frac { \tan ^ { 2 } A + 1 } { \tan ^ { 2 } A } } = \tan ^ { 2 } A \ldots \ldots ( 1 )$
$\left( \frac { 1 - \tan A } { 1 - \cot A } \right) ^ { 2 } = \left( \frac { 1 - \tan A } { 1 - \frac { 1 } { \tan A } } \right) ^ { 2 }$
$= \left\{ \frac { 1 - \tan A } { \left( \frac { \tan A - 1 } { \tan A } \right) } \right\} ^ { 2 } = ( - \tan A ) ^ { 2 } = \tan ^ { 2 } A$ $....... (2)$
$(1)$ and $(2)$ taken together given the result.
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Question 51 Mark
Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $( cosec\; \theta - \cot \theta ) ^ { 2 } = \frac { 1 - \cos \theta } { 1 + \cos \theta }$
Answer
$L H S = ( cosec\; \theta - \cot \theta ) ^ { 2 }$
$= \left( \frac { 1 } { \sin \theta } - \frac { \cos \theta } { \sin \theta } \right) ^ { 2 } = \left( \frac { 1 - \cos \theta } { \sin \theta } \right) ^ { 2 } = \frac { ( 1 - \cos \theta ) ^ { 2 } } { \sin ^ { 2 } \theta }$
$= \frac { ( 1 - \cos \theta ) ^ { 2 } } { 1 - \cos ^ { 2 } \theta } = \frac { ( 1 - \cos \theta ) ^ { 2 } } { ( 1 - \cos \theta ) ( 1 + \cos \theta ) }$
$= \frac { 1 - \cos \theta } { 1 + \cos \theta } = R H S$
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Question 111 Mark
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$ =
Answer
We have $\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$= $\frac{2\times\frac{1} {\sqrt{3}}}{1- (\frac{1} {\sqrt{3}})^{2} }$
= ${\frac2{\sqrt{3}}}\times{}{\frac3{{2}}}$
= ${\frac3{{\sqrt{3}}}}= \sqrt{3} $ = $\tan 60^{\circ}$
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Question 121 Mark
$sin\ 2A = 2\ sin\  A$ is true when $A =$
Answer
$sin\ 2A = 2\ sin\  A$ is true when A =${0^ \circ }$
$\because $$\sin {\text{2A}} = 2\sin {\text{A}}$
$ \Rightarrow $$\sin \left( {2 \times {0^ \circ }} \right) = \sin {0^ \circ }$
$ \Rightarrow $$\sin {0^ \circ } = \sin {0^ \circ }$
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Question 131 Mark
$\frac{{1 - {{\tan }^2}45^\circ }}{{1 + {{\tan }^2}45^\circ }} = $
Answer
Given: $\frac{{1 - {{\tan }^2}45^\circ }}{{1 + {{\tan }^2}45^\circ }}$
= $\frac{{1 - {{\left( 1 \right)}^2}}}{{1 + {{\left( 1 \right)}^2}}} = \frac{{1 - 1}}{{1 + 1}}$$= 0/2 = 0$
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Question 141 Mark
$\frac{{2\tan 30^\circ }}{{1 + {{\tan }^2}30^\circ }} = $
Answer
Given: $\frac{{2\tan 30^\circ }}{{1 + {{\tan }^2}30^\circ }}$
= $\frac{{2 \times \frac{1}{{\sqrt 3 }}}}{{1 + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}$
= $\frac{2}{{\sqrt 3 \left( {\frac{{3 + 1}}{3}} \right)}}$
= $\frac{6}{{4\sqrt 3 }} = \frac{3}{{2\sqrt 3 }}$
= $\frac{{\sqrt 3 }}{2}$
= $\sin {60^ \circ }$
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Question 151 Mark
Evaluate: $2 \tan ^2 45^{\circ}+\cos ^2 30^{\circ}-\sin ^2 60^{\circ}$
Answer
We have $2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$
now put values, we get
$ = 2{(1)^2} + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\frac{{\sqrt 3 }}{2}} \right)^2}$
$ = 2 + \frac{3}{4} - \frac{3}{4}$
= 2
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Question 161 Mark
Evaluate: $\sin\ 60^\circ \cos\ 30^\circ + \sin 30^\circ\ \cos 60^\circ $
Answer
$\sin 60 ^ { \circ } \cos 30 ^ { \circ } + \sin 30 ^ { \circ } \cos 60 ^ { \circ }$
$= \frac { \sqrt { 3 } } { 2 } \cdot \frac { \sqrt { 3 } } { 2 } + \frac { 1 } { 2 } \cdot \frac { 1 } { 2 }$
$= \frac { 3 } { 4 } + \frac { 1 } { 4 } = 1$
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Question 171 Mark
If $\angle A$ and $\angle B$ are acute angles such that $cos\ A = cos\ B$ then show that $\angle A = \angle B$.
Answer
In right triangle $ABC,$
$\cos A = \frac { A C } { A B } \ and \cos B = \frac { B C } { A B }$
But $cos\ A = cos\ B$ [Given]
$\Rightarrow \frac { A C } { A B } = \frac { B C } { A B } \Rightarrow _ { A C } = B C$
$\Rightarrow \angle A = \angle B$
[Angles opposite to equal sides are equal]
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Question 181 Mark
Evaluate $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$
Answer
We know : $\cot A = \tan (90^\circ – A)$
So, $\cot 25^\circ = \tan (90^\circ – 25^\circ ) = \tan 65^\circ$
i.e., $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}=\frac{\tan 65^{\circ}}{\tan 65^{\circ}}=1$
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Question 191 Mark
If sin $(A - B) =$ $\frac { 1 } { 2 }$, $cos (A + B) =$ $\frac { 1 } { 2 }$, $0^\circ < A + B$ $\leq$ $90^\circ , A > B$ find $A$ and $B$.
Answer
According to the question, $\sin (A - B) = \frac{1}{2}$, therefore, $A - B = 30^\circ........ (1)$
$\cos (A + B) = \frac{1}{2}$, therefore, $A + B = 60^\circ........ (2)$
Solving $(1)$ and $(2)$, we get
$A = 45^\circ$ and $B = 15^\circ$
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Question 201 Mark
Express $cot\ 85^\circ + \cos\ 75^\circ $ in terms of trigonometric ratios of angles between $0^\circ $ and $45^\circ $.
Answer
$\cot 85^\circ + \cos 75^\circ $
$= \tan (90^\circ - 85^\circ ) + \sin (90^\circ - 75^\circ )$
$= \tan 5^\circ + \sin 15$
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