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Maths 2 Marks Questions

Introduction to Trigonometry · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 23 questions.

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Question 12 Marks
Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $\frac { \sin \theta - 2 \sin ^ { 3 } \theta } { 2 \cos ^ { 2 } \theta - \cos \theta } = \tan \theta$
Answer
$LHS$
$= \frac { \sin \theta - 2 \sin ^ { 3 } \theta } { 2 \cos ^ { 2 } \theta - \cos \theta } = \frac { \sin \theta \left( 1 - 2 \sin ^ { 2 } \theta \right) } { \cos \theta \left( 2 \cos ^ { 2 } \theta - 1 \right) }$
$= \frac { \sin \theta \left( \cos ^ { 2 } \theta + \sin ^ { 2 } \theta - 2 \sin ^ { 2 } \theta \right) } { \cos \theta \left( 2 \cos ^ { 2 } \theta - \cos ^ { 2 } \theta - \sin ^ { 2 } \theta \right) } \quad \because \cos ^ { 2 } \theta + \sin ^ { 2 } \theta = 1$
$= \frac { \sin \theta \left( \cos ^ { 2 } \theta - \sin ^ { 2 } \theta \right) } { \cos \theta \left( \cos ^ { 2 } \theta - \sin ^ { 2 } \theta \right) } = \tan \theta$
$= RHS$
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Question 22 Marks
Prove the given identity, where the angles involved are acute angles for which the expressions are defined.
$\frac { \cos A } { 1 + \sin A } + \frac { 1 + \sin A } { \cos A } = 2 \sec A$
Answer
$L H S = \frac { \cos A } { 1 + \sin A } + \frac { 1 + \sin A } { \cos A }$
$= \frac { \cos ^ { 2 } A + ( 1 + \sin A ) ^ { 2 } } { ( 1 + \sin A ) \cos A } = \frac { \cos ^ { 2 } A + 1 + \sin ^ { 2 } A + 2 \sin A } { ( 1 + \sin A ) \cos A }$
$= \frac { 1 + 1 + 2 \sin A } { ( 1 + \sin A ) \cos A } \because \sin ^ { 2 } A + \cos ^ { 2 } A = 1$
$= \frac { 2 + 2 \sin A } { ( 1 + \sin A ) \cos A } = \frac { 2 ( 1 + \sin A ) } { ( 1 + \sin A ) \cos A }$
$= \frac { 2 } { \cos A } = 2 \cdot \frac { 1 } { \cos A } = 2 \sec A = R H S$
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Question 32 Marks
If $\tan (A + B) =$ $\sqrt3$ and $\tan (A - B) =$ $\frac{1}{\sqrt3}$; $0^\circ < A + B$ $\leq$ $90^\circ ; A > B$, then find $A$ and $B$.
Answer
We have, $\tan (A + B) =$ $\sqrt3$ $\Rightarrow$$\tan(A+B) = \tan 60^\circ$
$A + B = 60^\circ ..........(i)$
Again, $\tan (A - B )$ =$\frac{1}{\sqrt3}$ $\Rightarrow$$\tan(A-B) = \tan 30^\circ$
$A - B = 30^\circ ..........(ii)$
Adding, $(i)$ and $(ii)$
$2A = 90^\circ $
$\therefore$ A = $\frac{90^\circ}{2}$ $=45^\circ$
Putting $A=45$$^o$ in equation $(i)$,
$B = 60^\circ - A = 60^\circ - 45^\circ = 15^\circ $
Therefore, $A = 45^\circ$ and $B = 15^\circ$.
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Question 42 Marks
Evaluate: $\frac { 5 \cos ^ { 2 } 60 ^ { \circ } + 4 \sec ^ { 2 } 30 ^ { \circ } - \tan ^ { 2 } 45 ^ { \circ } } { \sin ^ { 2 } 30 ^ { \circ } + \cos ^ { 2 } 30 ^ { \circ } }$
Answer
Given: $\frac { 5 \cos ^ { 2 } 60 ^ { \circ } + 4 \sec ^ { 2 } 30 ^ { \circ } - \tan ^ { 2 } 45 ^ { \circ } } { \sin ^ { 2 } 30 ^ { \circ } + \cos ^ { 2 } 30 ^ { \circ } }$
$= \frac { 5 \left( \frac { 1 } { 2 } \right) ^ { 2 } + 4 \left( \frac { 2 } { \sqrt { 3 } } \right) ^ { 2 } - ( 1 ) ^ { 2 } } { \left( \frac { 1 } { 2 } \right) ^ { 2 } + \left( \frac { \sqrt { 3 } } { 2 } \right) ^ { 2 } }$
$= \frac { 5 \times \frac { 1 } { 4 } + 4 \times \frac { 4 } { 3 } - 1 } { \frac { 1 } { 4 } + \frac { 3 } { 4 } }$
$= \frac { \frac { 5 } { 4 } + \frac { 16 } { 3 } - 1 } { \frac { 1 + 3 } { 4 } }$
$= \frac { \frac { 15 + 64 - 12 } { 12 } } { \frac { 4 } { 4 } }$
$= \frac { \frac { 67 } { 12 } } { 1 }$
$= \frac { 67 } { 12 }$
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Question 52 Marks
Evaluate: $\frac{\sin 30^{\circ}+\tan 45^{\circ}-\ cosec 60^{\circ} }{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
Answer
We have $\frac{\sin 30^{\circ}+\tan 45^{\circ}- \ cosec 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
after putting values,we get
$=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}$
$=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}$
$=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4}$ Rationalise it, we get
$=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4} \times \frac{3 \sqrt{3}-4}{3 \sqrt{3}-4}$
$=\frac{(3 \sqrt{3}-4)^{2}}{(3 \sqrt{3})^{2}-(4)^{2}}$
$=\frac{27+16-24 \sqrt{3}}{27-16}$
$=\frac{43-24 \sqrt{3}}{11}$
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Question 62 Marks
Evaluate: $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+ \ cosec 30^{\circ}}$
Answer
We have $\frac{\cos \left(45^{\circ}\right)}{\sec \left(30^{\circ}\right)+\ cosec \left(30^{\circ}\right)}$
$=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}$
$=\frac{\frac{1}{\sqrt{2}}}{2\left(\frac{1}{\sqrt{3}}+1\right)}$
$=\frac{1}{2 \sqrt{2}\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)}$
$=\frac{\sqrt{3}}{2 \sqrt{2}(1+\sqrt{3})}$
it is clear that the denominator has an irrational number, we need to rationalize it, we get
$=\frac{\sqrt{3}}{2 \sqrt{2}(1+\sqrt{3})} \times \frac{\sqrt{2}(1-\sqrt{3})}{\sqrt{2}(1-\sqrt{3})}$
$=\frac{\sqrt{2}(\sqrt{3}-3)}{2(2)\left(1^{2}-(\sqrt{3})^{2}\right)}$
$=\frac{\sqrt{6}-3 \sqrt{2}}{4(1-3)}$
$=\frac{\sqrt{6}-3 \sqrt{2}}{4(-2)}$
$=\frac{\sqrt{6}-3 \sqrt{2}}{-8}$
$=\frac{3 \sqrt{2}-\sqrt{6}}{8}$
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Question 72 Marks
If $\cot$ $\theta$ $= \frac { 7 } { 8 }$, evaluate $\frac { ( 1 + \sin \theta ) ( 1 - \sin \theta ) } { ( 1 + \cos \theta ) ( 1 - \cos \theta ) }$.
Answer
Given: $\cot$ $\theta$ $= \frac { 7 } { 8 }$
To Evaluate: $\frac { ( 1 + \sin \theta ) ( 1 - \sin \theta ) } { ( 1 + \cos \theta ) ( 1 - \cos \theta ) }$
$= \frac { 1 - \sin ^ { 2 } \theta } { 1 - \cos ^ { 2 } \theta } = \frac { \cos ^ { 2 } \theta } { \sin ^ { 2 } \theta }$
$= cot^2\theta$
$= \left( \frac { 7 } { 8 } \right) ^ { 2 }$
$= \frac { 49 } { 64 }$
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Question 82 Marks
If $\sin A = \frac { 3 } { 4 }$, calculate $\cos A$ and $\tan A$.
Answer
Given: $A$ triangle $ABC$ in which $\angle B = 90 ^ { \circ }$

$SinA=\frac34=\frac PH$
Let $BC = 3k$ and $AC = 4k$ where $k$ is a positive integer.$|$
Using Pythagoras theorem,
$AB^2=AC^2-BC^2$
$A B = \sqrt { ( A C ) ^ { 2 } - ( B C ) ^ { 2 } } = \sqrt { ( 4 k ) ^ { 2 } - ( 3 k ) ^ { 2 } }$
$= \sqrt { 16 k ^ { 2 } - 9 k ^ { 2 } } =\sqrt{7k^2}= k \sqrt { 7 }$
Therefore,$\cos A = \frac { B } { H } = \frac { A B } { A C } = \frac { k \sqrt { 7 } } { 4 k } = \frac { \sqrt { 7 } } { 4 }$
$\tan \mathrm { A } = \frac { \mathrm { P } } { \mathrm { B } } = \frac { \mathrm { BC } } { \mathrm { AB } } = \frac { 3 k } { k \sqrt { 7 } } = \frac { 3 } { \sqrt { 7 } }$
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Question 92 Marks
In figure, find $\tan P - \cot R$.
Answer
In $\triangle P Q R$ ,
$\therefore \angle Q = 90 ^ { \circ }$
$ \therefore P R^2=P Q^2+Q R^2 \ldots \ldots . . \text { By Pythagoras theorem } $
$ \Rightarrow(13)^2=(12)^2+Q R^2 $
$ \Rightarrow 169=144+Q R^2 $
$ \Rightarrow Q R^2=169-144 \Rightarrow Q R^2=25$
$\Rightarrow Q R = \sqrt { 25 } = 5 \mathrm { cm }$
$\therefore \tan \mathrm { P } - \cot \mathrm { R } = \frac { \mathrm { QR } } { \mathrm { PQ } } - \frac { \mathrm { QR } } { \mathrm { PQ } } = 0$
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Question 152 Marks
In $\triangle$$PQR$, right-angled at $Q, PQ = 3\ cm$ and $PR = 6\ cm$. Determine $\angle$$QPR$ and $\angle$$PRQ$.
Answer
We have given that $PQ = 3\ cm$ and $PR = 6\ cm$
from $\triangle$$PQR$, $\frac{P Q}{P R}=\sin R$
or $\sin R=\frac{3}{6}=\frac{1}{2}$
So, $\angle$$PRQ = 30^\circ$
and therefore, $\angle$$QPR = 60^\circ$ (by using angle sum property of triangle)
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Question 162 Marks
Prove that: $\frac{\cot A-\cos A}{\cot A+\cos A}$$=\frac{\ cosec A-1}{\ cosec A+1}$
Answer
$LHS$ = $\frac{\cot A-\cos A}{\cot A+\cos A}$
$=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}$
$=\frac{\frac{\cos A-\sin A \cos A}{\sin A}}{\frac{\cos A+\sin A \cos A}{\sin A}}$
$=\frac{\cos A(1-\sin A)}{\cos A(1+\sin A)}$
$=\frac{1-\sin A}{1+\sin A}$
$=\frac{\frac{1}{\sin A}-1}{\frac{1}{\sin A}+1}$
$=\frac{\ cosec A-1}{\ cosec A+1}$ $= RHS$
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Question 172 Marks
Prove that: $sec\ A (1 - \sin A) (sec\ A + \tan\ A) = 1$
Answer
$L.H.S.$ $= sec A(1 - \sin A)(sec A + \tan A)$
=$\frac{1}{cos A}$$(1 - sin A)$( $\frac{1}{cos A}$ + $\frac{sin A}{cos A}$)
$=\frac{(1 - sin A)}{cos A}$($\frac{1 + sin A}{cos A}$)
$= \frac{(1-\sin A)(1+\sin A)}{\cos A \times \cos A}$
$=\frac{\left(1^{2}-\sin ^{2} A\right)}{\cos ^{2} A} .[ $ Since, $(a - b ) (a + b ) = a^2- b^2]​​$
$ = \frac{\left(1-\sin ^{2} A\right)}{\cos ^{2} A}$ = $\frac{cos^2 A}{cos^2A}$
$= 1$
$=RHS$
Hence, proved.
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Question 182 Marks
Express the ratios cos A, tan A and sec A in terms of sin A.
Answer
As we know that $\cos ^2 \mathrm{~A}+\sin ^2 \mathrm{~A}=1$
therefore $\cos ^2 \mathrm{~A}=1-\sin ^2 \mathrm{~A}$
This gives $\cos A=\sqrt{1-\sin ^{2} A}$
Hence,$\tan A=\frac{\sin A}{\cos A}=\frac{\sin A}{\sqrt{1-\sin ^{2} A}} $
and $\sec A=\frac{1}{\cos A}=\frac{1}{\sqrt{1-\sin ^{2} A}}$
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Question 192 Marks
If $\sin 3 A = \cos \left( A - 26 ^ { \circ } \right)$, where $3A$ is an acute angle, find the value of $A$.
Answer
We are given that sin $3A = cos (A - 26^\circ) ...... (1)$
Since $\sin 3 A=\cos \left(90^{\circ}-3 A\right)$, we can write $(1)$ as
$\cos \left(90^{\circ}-3 A\right)=\cos \left(A-26^{\circ}\right)$
Since $90^{\circ}-3 A$ and $A-26^{\circ}$ are both acute angles, therefore,
$ 90^{\circ}-3 A=A-26^{\circ}$
$ 4 A=116^{\circ} $
$ \Rightarrow A=29^{\circ}$
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Question 202 Marks
Evaluate :$2 \tan ^2 45^{\circ}+\cos ^2 30^{\circ}-\sin ^2 60^{\circ}$
Answer
$\begin{aligned} 2 \tan ^2 45^{\circ}+\cos ^2 30^{\circ}-\sin ^2 60^{\circ}
& =2(1)^2+\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2
\\ & =2+\frac{3}{4}-\frac{3}{4} \\
& =2\end{aligned}$
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Question 212 Marks
If $\sin \theta=\frac{3}{5}$ find the value of cos$\theta$ and tan$\theta$
Answer
Let us take right angled triangle $A B C$ where $\angle B=90^{\circ}$ and $\angle C=\theta$.
Then, $\sin \theta=\frac{3}{5}=\frac{A B}{A C}$
$\therefore A C=5 k, A B=3 k$; (where, $k$ is some positive integer.)
$\therefore$ By Pythagoras theorem,
$
\begin{aligned}
B C & =\sqrt{A C^2-A B^2} \\
& =\sqrt{(5 k)^2-(3 k)^2} \\
& =\sqrt{25 k^2-9 k^2}=\sqrt{16 k^2} \\
\therefore B C & =4 k
\end{aligned}
$
Now, $\cos \theta=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$
and $\tan \theta=\frac{A B}{B C}=\frac{3 k}{4 k}=\frac{3}{4}$0
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Question 222 Marks
Find the value: \begin{equation}4 \cot ^2 45^{\circ}-\sec ^2 60^{\circ}+\sin ^2 60^{\circ}+\cos ^2 90^{\circ}\end{equation}
Answer
Given expression:
$4 \cot ^2 45^{\circ}-\sec ^2 60^{\circ}+\sin ^2 60^{\circ}+\cos ^2 90^{\circ}$
We know the values:
$\begin{array}{l}\cot 45^{\circ}=1 \\ \sec 60^{\circ}=2 \\ \sin 60^{\circ}=\frac{\sqrt{3}}{2} \\ \cos 90^{\circ}=0\end{array}$
Substitute these values into the expression:
$=4(1)^2-(2)^2+\left(\frac{\sqrt{3}}{2}\right)^2+(0)^2$
Now, perform the calculations:
$\begin{array}{l}=4(1)-4+\frac{3}{4}+0 \\ =4-4+\frac{3}{4} \\ =0+\frac{3}{4} \\ =\frac{3}{4}\end{array}$
The value of the expression is $\frac{3}{4}$
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Question 232 Marks
Prove that $\cos ^2 \theta-\sin ^2 \theta=2 \cos ^2 \theta-1$
Answer
$LHS =\cos ^2 \theta-\sin ^2 \theta$
From the fundamental trigonometric identity, we know that $\sin ^2 \theta+\cos ^2 \theta=1$.
This can be rewritten as $\sin ^2 \theta=1-\cos ^2 \theta$
Substitute this expression for $\sin ^2 \theta$ into the LHS:
$\begin{array}{l}\text { LHS }=\cos ^2 \theta-\left(1-\cos ^2 \theta\right) \\ \text { LHS }=\cos ^2 \theta-1+\cos ^2 \theta \\ \text { LHS }=2 \cos ^2 \theta-1\end{array}$
Since LHS = RHS, the identity is proven.
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