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Maths 2 Marks Questions

Introduction of Trigonometry · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 24 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.$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$ = cosec A + cot A, using the identity $cosec^2 A = 1 + cot^2 A$
Answer
Taking L.H.S
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$
Dividing Numerator and Denominator by sin A
$=\frac{\frac{\cos A-\sin A+1}{\sin A}}{\frac{\cos A+\sin A-1}{\sin A}}$
$=\frac{\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}+\frac{1}{\sin A}}{\frac{\cos A}{\sin A}+\frac{\sin A}{\sin A}-\frac{1}{\sin A}}$
Using the formula $\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}$$=\frac{\cot A-1+\ cosec A}{\cot A+1-\ cosec A}$
Using the identity $cosec^2A = 1 + cot^2A$
$=\frac{\cot A-\left(\ cosec ^{2} A-\cot ^{2} A\right)+\ cosec A}{\cot A+1-\ cosec A}$
$=\frac{(\cot A+\ cosec A)-\left(\ cosec ^{2} A-\cot ^{2} A\right)}{\cot A+1-\ cosec A}$
$=\frac{(\cot A+\ cosec A)(1-\ cosec A+\cot A)}{\cot A+1-\ cosec A}$
$= cot A + cosec A$
$= R.H.S$
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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 42 Marks
If tan (A + B) = $\sqrt3$ and tan (A - B) = $\frac{1}{\sqrt3}$; 0° < A + B $\leq$ 90°; A > B, then find A and B.
Answer
We have,

tan (A + B) = $\sqrt3$

$\Rightarrow$tan(A+B) = tan 60°
A + B = 60°..........(i)
Again, tan (A - B ) =$\frac{1}{\sqrt3}$

$\Rightarrow$tan(A-B) = tan 30°
A - B = 30°..........(ii)
Adding, (i) and (ii)
2A = 90°
$\therefore$ A = $\frac{90^\circ}{2}$ =45°
Putting A=45$^o$ in equation (i),
B = 60° - A = 60° - 45° = 15°
Therefore,

A = 45° and B = 15°.

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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
Given $\sec \theta = \frac { 13 } { 12 }$, Calculate all other trigonometric ratios.
Answer
Give that 3cot A = 4
Or cot $A = \frac{4}{3}$
Consider a right angle triangle $\Delta ABC$ right angled at point B.

$\cot A = \frac{{Side\;adjacent\;to\;\angle A}}{{Side\;opposite\;to\;\angle A}}$
$\frac{{AB}}{{BC}} = \frac{4}{3}$
If AB is 4K, BC will be 3K. where K is a positive integer
Now in $\Delta ABC$
$(AC)^2 = (AB)^2 + (BC)^2$
$= (4K)^2 + (3K)^2$
$= 16 K^2 + 9K^2$
$= 25K^2$
$AC = 5K$
$\cos A = \frac{{Side\;adjacent\;to\;\angle A}}{{hypotenuse}} = \frac{{AB}}{{AC}}$
$ = \frac{{4K}}{{5K}} = \frac{4}{5}$
$\sin A = \frac{{Side\;opposite\;to\;\angle A}}{{hypotenuse}} = \frac{{BC}}{{AC}}$
$ = \frac{{3K}}{{5K}} = \frac{3}{5}$
$\tan A = \frac{{Side\;opposite\;to\;\angle A}}{{Side\;adjacent\;to\;angle\;A}} = \frac{{BC}}{{AB}}$
$ = \frac{{3K}}{{4K}} = \frac{3}{4}$
$\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = \frac{{1 - {{\left( {\frac{3}{4}} \right)}^2}}}{{1 + {{\left( {\frac{3}{4}} \right)}^2}}} = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}}$
$ = \frac{{\frac{7}{{16}}}}{{\frac{{25}}{{16}}}} = \frac{7}{{25}}$
${\cos ^2}A - {\sin ^2}A = {\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2}$
$ = \frac{{16}}{{25}} - \frac{9}{{25}} = \frac{7}{{25}}$
Hence $\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A$
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Question 92 Marks
If $\sin A = \frac { 3 } { 4 }$, calculate cos A and tan A.
Answer

Given: A triangle ABC in which $\angle B = 90 ^ { \circ }$
Image
$SinA=\frac34=\frac PH$
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 102 Marks
In figure, find tan P - cot R.

Answer
Let us draw a right triangle ABC.
15 cot A = 8 ...... Given
$\Rightarrow \cot A = \frac { 8 } { 15 } \Rightarrow \frac { A B } { B C } = \frac { 8 } { 15 }$
$\Rightarrow \frac { A B } { 8 } = \frac { B C } { 15 } = k ( 5 a y )$
where k is a positive number
$\Rightarrow A B = 8 k$
BC = 15k

By using the Pythagoras theorem, we have
$A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 }$
$\Rightarrow A C ^ { 2 } = ( 8 k ) ^ { 2 } + ( 15 k ) ^ { 2 } \Rightarrow A C ^ { 2 } = 64 k ^ { 2 } + 225 k ^ { 2 }$
$\Rightarrow A C = \sqrt { 289 k ^ { 2 } } \Rightarrow A C = 17 k$
Now, $\sin A = \frac { B C } { A C } = \frac { 15 k } { 17 k } = \frac { 15 } { 17 }$
and, $\sec A = \frac { A C } { A B } = \frac { 17 k } { 8 k } = \frac { 17 } { 8 }$
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Question 212 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°
and therefore, $\angle$QPR = 60° (by using angle sum property of triangle)
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Question 222 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 232 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 242 Marks
Express the ratios $\cos A, \tan A $and $\sec A$ in terms of $\sin A.$
Answer
As we know that $\cos^2A + \sin^2A = 1$
therefore $\cos^2A = 1 - \sin^2A$
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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