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Published on: 13/05/2022
QB365 provides detailed and simple solution for every book back questions in class 10 Maths subject.It will helps to get more idea about question pattern in every book back questions with solution.
latest Book back Questions10роЖроорпН ро╡роХрпБрокрпНрокрпБ Maths / роХрогро┐родроорпН родрооро┐ро┤рпН роорпАроЯро┐ропроорпН | рооро╛родро┐ро░ро┐ ро╡ро┐ройро╛родрпНродро╛ро│рпНроХро│рпН | роорпБроХрпНроХро┐роп ро╡ро┐ройро╛роХрпНроХро│рпН | роХрпЗро│рпНро╡ро┐ ро╡ро┐роЯрпИ родрпКроХрпБрокрпНрокрпБ | ро╡ро┐ройро╛родрпНродро╛ро│рпНроХро│рпН | рокрпКродрпБродрпНродрпЗро░рпНро╡рпБ рооро╛родро┐ро░ро┐ ро╡ро┐ройро╛родрпНродро╛ро│рпНроХро│рпН , Maths / роХрогро┐родроорпН рокро╛роЯродрпНродро┐ро▒рпНроХро╛рой рооро╛родро┐ро░ро┐ ро╡ро┐ройро╛родрпНродро╛ро│рпНроХро│рпН, роорпБроХрпНроХро┐роп роХрпЗро│рпНро╡ро┐-ро╡ро┐роЯрпИроХро│рпН, роТро░рпБрооро╛ро░рпНроХрпН роХрпЗро│рпНро╡ро┐роХро│рпН, роХро╛ро▓ро╛рогрпНроЯрпБ & рокрпКродрпБродрпНродрпЗро░рпНро╡рпБ роХрпЗро│рпНро╡ро┐родрпН родрпКроХрпБрокрпНрокрпБроХро│рпН
Questions + Answers key
Take MCQ Maths Test1.
tan2 A - tan2 B \(=\frac { { \sin }^{ 2 }A-{ \sin }^{ 2 }B }{ { \cos }^{ 2 }A{ \cos }^{ 2 }B } \) роОройрпНрокродрпИ роиро┐ро░рпВрокро┐роХрпНроХро╡рпБроорпН.
2.
(cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ) = 1 роОройрпНрокродрпИ роиро┐ро░рпВрокро┐роХрпНроХро╡рпБроорпН.
3.
\(\frac { (1+\cot A+\tan A)(\sin A- \cos A) }{ { \sec }^{ 3 }A-{ cosec }^{ 3 }A } ={ \sin }^{ 2 }A{ \cos }^{ 2 }A\) роОройрпНрокродрпИ роиро┐ро░рпВрокро┐роХрпНроХро╡рпБроорпН.
4.
\(\frac { \sin A }{ \sec A+\tan A-1 } +\frac { \cos A }{ cosecA+\cot A-1 } =1\) роОройрпНрокродрпИ роиро┐ро░рпВрокро┐роХрпНроХро╡рпБроорпН.
5.
cosec θ + cot θ = P роОройро┐ро▓рпН, cos θ = \(\frac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 } \) роОройрпНрокродрпИ роиро┐ро░рпВрокро┐роХрпНроХро╡рпБроорпН.
1.
tan2 A - tan2 B = \(\frac { { \sin }^{ 2 }A }{ { \cos }^{ 2 }A } -\frac { { \sin }^{ 2 }B }{ { \cos }^{ 2 }B } \)
\(=\frac { { \sin }^{ 2 }A{ \cos }^{ 2 }B-{ \sin }^{ 2 }B{ \cos }^{ 2 }A }{ { \cos }^{ 2 }A{ \cos }^{ 2 }B } \)
\(=\frac{\sin ^2 A\left(1-\sin ^2 B\right)-\sin ^2 B\left(1-\sin ^2 A\right)}{\cos ^2 A \cos ^2 B}\)
\(=\frac { { \sin }^{ 2 }A-{ \sin }^{ 2 }A{ \sin }^{ 2 }B-{ \sin }^{ 2 }B+{ \sin }^{ 2 }A{ \sin }^{ 2 }B }{ { \cos }^{ 2 }A{ \cos }^{ 2 }B } =\frac { { \sin }^{ 2 }A-{ \sin }^{ 2 }B }{ { \cos }^{ 2 }A{ \cos }^{ 2 }b } \)
2.
(cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ)
\(=\left( \frac { 1 }{ \sin\theta } -\sin \theta \right) \left( \frac { 1 }{ \cos\theta } -\cos\theta \right) \left( \frac { \sin\theta }{ \cos\theta } +\frac { \cos\theta }{ \sin\theta } \right) \)
\(=\frac { 1-{ \sin }^{ 2 }\theta }{ \sin\theta } \times \frac { 1-{ \cos }^{ 2 }\theta }{ \cos\theta } \times \frac { { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta }{ \sin\theta \cos\theta } \)
\(=\frac { { \cos }^{ 2 }\theta { \sin }^{ 2 }\theta \times 1 }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta } =1\)
3.
\(\frac { (1+\cot A+\tan A)(\sin A-\cos A) }{ { \sec }^{ 3 }A-{ cosec }^{ 3 }A } \)
= \(\frac { \left( 1+\frac { \cos A }{ \sin A } +\frac { \sin A }{ \cos A } \right) \left( \sin A-\cos A \right) }{ (\sec A-cosecA)({ \sec }^{ 2 }A+\sec A cosecA+{ cosec }^{ 2 }A } \)
= \(\frac { \frac { (\sin A \cos A+{ \cos }^{ 2 }A+{ \sin }^{ 2 }A)(\sin A-\cos A) }{ \sin A \cos A } }{ (\sec A- cosecA)\left( \frac { 1 }{ { \cos }^{ 2 }A } +\frac { 1 }{ \cos A \sin A } +\frac { 1 }{ { \sin }^{ 2 }A } \right) } \)
= \(\frac { (\sin A \cos A+1)\left( \frac { \sin A }{ \sin A\cos A } -\frac { \cos A }{ \sin A \cos A } \right) }{ (\sec A-cosecA)\left( \frac { { \sin }^{ 2 }A+\sin A \cos A+{ \cos }^{ 2 }A }{ { \sin }^{ 2 }A{ \cos }^{ 2 }A } \right) } \)
= \(\frac { (\sin A \cos A+1)(\sec A-cosecA) }{ (\sec A-cosecA)(1+\sin A \cos A) } \times { \sin }^{ 2 }A{ \cos }^{ 2 }A={ \sin }^{ 2 }A{ \cos }^2A\)
4.
\(\frac { \sin A }{ \sec A+\tan A-1 } +\frac { \cos A }{ cosecA+\cot A-1 } =1\)
= \(\frac { \sin A(cosecA+\cot A-1)+\cos A(\sec A+\tan A-1) }{ (\sec A+\tan A-1)(cosec A+\cot A-1) } \)
= \(\frac { \sin A cosecA+\sin A \cot A-\sin A+\cos A \sec A+\cos A\tan A-\cos A }{ (\sec A+\tan A-1)(cosecA+\cot A-1) } \)
= \(\frac { 1+\cos A-\sin A+1+\sin A-\cos A }{ \left( \frac { 1 }{ \cos A } +\frac { \sin A }{ \cos A } -1 \right) \left( \frac { 1 }{ \sin A } +\frac { \cos A }{ \sin A } -1 \right) } \)
= \(\frac { 2 }{ \left( \frac { 1+sinA-cosA }{ cosA } \right) \left( \frac { 1+cosA-sinA }{ sinA } \right) } \)
= \(\frac { 2sinAcosA }{ (1+sinA-cosA)(1+cosA-sinA) } \)
= \(\frac { 2sinAcosA }{ [1+(sinA-cosA)][1-(sinA-cosA)] } =\frac { 2sinAcosA }{ 1-{ (sinA-cosA) }^{ 2 } } \)
= \(\frac { 2sinAcosA }{ 1-({ sin }^{ 2 }A+{ cos }^{ 2 }A-2sinAcosA) } =\frac { 2sinAcosA }{ 1-(1-2sinAcosA) } \)
= \(\frac { 2sinAcosA }{ 1-(1+2sinAcosA) } \) = \(\frac { 2sinAcosA }{ 2sinAcosA} \) = 1
5.
\(\operatorname{cosec} \theta+\cot \theta=P\) роОройроХрпН роХрпКроЯрпБроХрпНроХрокрпНрокроЯрпНроЯрпБро│рпНро│родрпБ .....(1)
\(\operatorname{cosec}^2 \theta-\cot ^2 \theta =1\) (роорпБро▒рпНро▒рпКро░рпБроорпИ)
\(\operatorname{cosec} \theta-\cot \theta =\frac{1}{\operatorname{cosec} \theta+\cot \theta} \)
\(\operatorname{cosec} \theta-\cot \theta =\frac{1}{P}\) .....(2)
(1) рооро▒рпНро▒рпБроорпН (2) роЖроХро┐ропро╡ро▒рпНро▒рпИроХрпН роХрпВроЯрпНроЯроХрпНроХро┐роЯрпИрокрпНрокродрпБ, \(2 \operatorname{cosec} \theta=P+\frac{1}{P}\)
\(2 \operatorname{cosec} \theta=\frac{P^2+1}{P} \quad \ldots \text { (3) }\)
(2) -ро▓ро┐ро░рпБроирпНродрпБ (1) -роР роХро┤ро┐родрпНродро╛ро▓рпН роХро┐роЯрпИрокрпНрокродрпБ, \(2 \cot \theta=P-\frac{1}{P}\)
\(2 \cot \theta=\frac{P^2-1}{P} \quad \ldots(4)\)
(4) -роР (3) -роЖро▓рпН ро╡роХрпБроХрпНроХроХрпН роХро┐роЯрпИрокрпНрокродрпБ, \(\frac{2 \cot \theta}{2 \operatorname{cosec} \theta}=\frac{P^2-1}{P} \times \frac{P}{P^2+1}\)
\(\Rightarrow \cos \theta=\frac{P^2-1}{P^2+1}\) роОройроХрпН роХро┐роЯрпИроХрпНроХро┐ро▒родрпБ.
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