SSC/Class 9 10 math exercise 9.2(trigonometry) solution : part 2

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SSC/Class 9 10 math exercise 9.2(trigonometry) solution: part 2

7(i) . \frac{sinA}{cosecA} + \frac{cosA}{secA} = 1

বামপক্ষ = \frac{sinA}{cosecA} + \frac{cosA}{secA}
= sinA × \frac{1}{cosecA} + cosA × \frac{1}{secA}
= sinA × sinA + cosA × cosA
= sin^2A + cos^2A
= 1

= ডানপক্ষ

অতএব, \frac{sinA}{cosecA} + \frac{cosA}{secA} = 1 

(ii) \frac{secA}{cosA} - \frac{tanA}{cotA} = 1

বামপক্ষ = \frac{secA}{cosA} - \frac{tanA}{cotA}
= \frac{secA}{cosA} - \frac{tanA}{cotA}
= secA × \frac{1}{cosA} - tanA × \frac{1}{cotA}
= sec^2A - tan^2A
= 1

 \frac{secA}{cosA} - \frac{tanA}{cotA} = 1

 

(iii) \frac{1}{1 + sin^2A} +\frac{1}{1 + cosec^2A} = 1

বামপক্ষ = \frac{1}{1 + sin^2A} + \frac{1}{1 + cosec^2A}
= \frac{1}{1 + sin^2A} + \frac{1}{1 + \frac{1}{sin^2A}}
= \frac{1}{1 + sin^2A} + \frac{1}{\frac{1 + sin^2A}{sin^2A}}
= \frac{1}{1 + sin^2A} + \frac{sin^2A}{1 + sin^2A}
= \frac{1 + sin^2A}{1 + sin^2A}
= 1

 

8(i) \frac{tanA}{1 - cotA} + \frac{cotA}{1 - tanA} = secAcosecA + 1

বামপক্ষ = \frac{tanA}{1 - cotA} + \frac{cotA}{1 - tanA}

= \frac{\frac{sinA}{cosA}}{1 - \frac{cosA}{sinA}} + \frac{\frac{cosA}{sinA}}{1 - \frac{sinA}{cosA}}

= \frac{\frac{sinA}{cosA}}{\frac{sinA - cosA}{sinA}} + \frac{\frac{cosA}{sinA}}{\frac{cosA - sinA}{cosA}}

= \frac{sinA}{cosA} × (\frac{sinA}{ sinA - cosA}) + \frac{cosA}{sinA} × \frac{cosA}{ cosA - sinA}

= \frac{sin^2A}{cosA(sinA - cosA)} + \frac{cos^2A}{sinA(cosA - sinA)}

= \frac{sin^2A}{cosA(sinA - cosA)} - \frac{cos^2A}{sinA(sinA - cosA)}

= \frac{sin^3A - cos^3A}{sinAcosA(sinA - cosA)}

= \frac{(sinA - cosA)(sin^2A + sinAcosA + cos^2A)}{sinAcosA(sinA - cosA)}

= \frac{sin^2A + cos^2A + sinAcosA }{sinAcosA}

= \frac{1 + sinAcosA }{sinAcosA}

= \frac{1}{ sinAcosA} + \frac{sinAcosA}{sinAcosA}

= \frac{1}{sinA} . \frac{1}{cosA} + 1

= cosecA secA + 1

SSC/Class 9 10 math exercise 9.2(trigonometry) solution

(ii) \frac{1}{1 + \tan^2 A} + \frac{1}{1 + \cot^2 A} = 1

Step-by-Step Solution for SSC/Class 9-10 Math Exercise 9.2 (Trigonometry) Explained

বামপক্ষ = \frac{1}{1 + \tan^2 A} + \frac{1}{1 + \cot^2 A}

= \frac{1}{1 + \tan^2 A} + \frac{1}{1 + \frac{1}{\tan^2 A}}
= \frac{1}{1 + \tan^2 A} + \frac{\tan^2 A}{\tan^2 A + 1}
= \frac{1 + \tan^2 A}{1 + \tan^2 A}

= 1

= ডানপক্ষ

অতএব,  \frac{1}{1 + \tan^2 A} + \frac{1}{1 + \cot^2 A} = 1

 

৯। \frac{cos A}{1 - tan A} + \frac{sin A}{1 - cot A} = sin A + cos A

বামপক্ষ = \frac{cos A}{1 - tan A} + \frac{sin A}{1 - cot A}

= \frac{cos A}{1 - \frac{sinA}{cosA}} + \frac{sin A}{1 - \frac{cosA}{sinA}}

= \frac{cos A}{\frac{cosA - sinA}{cosA}} + \frac{sin A}{\frac{sinA - cosA}{sinA}}

= \frac{cos ^2A}{cosA - sinA} - \frac{sin^2A}{cosA - sinA}

= \frac{cos ^2A - sin^2A }{cosA - sinA}

= \frac{(cos A - sin A)(cos A + sin A)}{cos A - sin A}
= cos A + sin A 

= ডানপক্ষ

অতএব,  \frac{1}{1 - tan A} + \frac{1}{1 - cot A} = sin A + cos A

 

১০। \tan A \sqrt{1 - sin^2 A} = sin A

সমাধান :বামপক্ষ

= \tan A \sqrt{1 - \sin^2 A}

= \tan A \cos^2 A

= \frac{sin A}{cos A} × cos A

= sin A

= ডানপক্ষ

অতএব,  \tan A \sqrt{1 - \sin^2 A} = \sin A

Master SSC/Class 9-10 Math Exercise 9.2 (Trigonometry) with Easy Solutions

১১।  \frac{sec A + \tan A}{cosecA + cot A} = \frac{cosecA - cot A}{sec A - tan A}

সমাধান:

বামপক্ষ = \frac{sec A + tan A}{cosecA + cot A}

= \frac{(sec A + tan A)(sec A - tan A)}{(cosecA + cot A)(sec A - tan A)} \times \frac{(cosecA - cot A)}{(cosecA - cot A)}

[হর ও লবকে একই রাশি দ্বারা গুণ করে]

= \frac{(sec A + tan A)(sec A - tan A)}{(cosecA + cot A)(cosecA - cot A)} \times \frac{(cosecA - cot A)}{(sec A - tan A)}

= \frac{sec^2 A - tan^2 A}{cosec^2 A - cot^2 A} × \frac{cosecA - cot A}{sec A - tan A}

= \frac{1.(cosecA - cot A)}{1.(sec A - \tan A)} [\because sec^2 A - tan^2 A = 1 , cosec^2 A - cot^2 A = 1 ]

= \frac{cosecA - cot A}{sec A - tan A}

= ডানপক্ষ

অতএব,

\frac{sec A + tan A}{cosecA + cot A} = \frac{cosecA - cot A}{sec A - tan A}

প্রমাণিত

SSC/Class 9 10 math exercise 9.2(trigonometry) solution

১২। \frac{cosecA}{cosecA - 1} + \frac{cosecA}{cosecA + 1} = 2 sec^2 A

সমাধান:

বামপক্ষ = \frac{cosecA}{cosec A - 1} + \frac{\csc A}{cosec A + 1}

= \frac{cosecA(cosecA + 1) + cosec A (cosecA - 1)}{(cosecA - 1)(cosecA + 1)}

= \frac{cosec^2 A - cosecA + cosec^2 A + cosecA}{\cosec^2 A - 1}

= \frac{2 cosec^2 A}{1 + cot^2 A - 1} [\because cosec^2 A = 1 + cot^2 A ]

= \frac{2 cosec^2 A}{cot^2 A}

= \frac{\frac{2}{sin^2A}}{\frac{cos^2A}{sin^2A}}

= \frac{2}{sin^2 A} \times \frac{sin^2 A}{cos^2 A}

= \frac{2}{cos^2 A}

= 2 sec^2 A [\because sec A = \frac{1}{\cos A}]

= ডানপক্ষ

অতএব,

\frac{cosec A}{cosec A - 1} + \frac{cosec A}{cosec A + 1} = 2 sec^2 A (প্রমাণিত)

 

১৩। \frac{1}{1 + sinA} + \frac{1}{1 - sinA} = 2sec^2A

সমাধান:

বামপক্ষ = \frac{1}{1 + sinA} + \frac{1}{1 - sinA}

 

= \frac{1 - sinA + 1 +sinA}{(1 + sinA)( 1 - sinA)}

 

= \frac{2}{1 - sin^2A}

 

= \frac{2}{cos^2A} [\because 1 - sin^2A = cos^2A ]

 

= 2 sec^2A

= ডানপক্ষ

অতএব,

\frac{1}{1 + sinA} + \frac{1}{1 - sinA} = 2sec^2A (প্রমাণিত)

Complete Guide to SSC/Class 9-10 Math Exercise 9.2 (Trigonometry) Solutions

১৪। \frac{1}{cosecA - 1} - \frac{1}{ cosecA + 1} = 2tan^2A

সমাধান: 

বামপক্ষ = \frac{1}{cosecA - 1} - \frac{1}{ cosecA + 1}

 

= \frac{ cosecA + 1 - cosecA + 1}{(cosecA - 1)( cosecA + 1)}

 

= \frac{ 2}{cosec^2A - 1}

 

= \frac{ 2}{cot^2A} [\because cosec^2A - 1 = cot^2A]

 

= \frac{ 2}{cot^2A} [\because cosec^2A - 1 = cot^2A]

 

= 2 \times(\frac{ 1}{cotA})^2

 

= 2 tan^2A [\because \frac{1}{cotA} = tanA]   

= ডানপক্ষ

অতএব,

\frac{1}{cosecA - 1} - \frac{1}{ cosecA + 1} = 2tan^2A  (প্রমাণিত)

 

১৫। \frac{sinA}{1 - cosA} + \frac{1 - cosA}{sinA} = 2cosecA  

সমাধান: 

বামপক্ষ : \frac{sinA}{1 - cosA} + \frac{1 - cosA}{sinA}  

 

= \frac{sin^2A + (1 - cosA)^2}{sinA(1 - cosA)}

 

=  \frac{sin^2A + 1 - 2cosA + cos^2A}{sinA(1 - cosA)}

 

=  \frac{sin^2A + cos^2A + 1 - 2cosA }{sinA(1 - cosA)}

 

=  \frac{1 + 1 - 2cosA }{sinA(1 - cosA)}

 

= \frac{2 - 2cosA }{sinA(1 - cosA)}

 

= \frac{2(1 - cosA) }{sinA(1 - cosA)}

 

= \frac{2}{sinA}

 

= 2cosecA

= ডানপক্ষ

অতএব, \frac{sinA}{1 - cosA} + \frac{1 - cosA}{sinA} = 2cosecA  

Trigonometry Made Easy: SSC/Class 9-10 Math Exercise 9.2 Solution

১৬। \frac{tanA}{secA + 1} - \frac{secA - 1}{tanA} = 0

সমাধান: 

বামপক্ষ = \frac{tanA}{secA + 1} - \frac{secA - 1}{tanA}

 

= \frac{tan^2A - (secA + 1)( secA - 1)}{tanA (secA + 1)}

 

= \frac{tan^2A - ( sec^2A - 1)}{tanA (secA + 1)}

 

= \frac{tan^2A - tan^2A}{tanA (secA + 1)}

 

= \frac{0}{tanA (secA + 1)}

 

= 0

= ডানপক্ষ

অতএব, \frac{tanA}{secA + 1} - \frac{secA - 1}{tanA} = 0

 

১৭। (tanθ + secθ)^2 = \frac{1 + sinθ}{1 - sinθ}

সমাধান: 

বামপক্ষ = (tanθ + secθ)^2

 

=  (\frac{sinθ}{cosθ} + \frac{1}{cosθ})^2

 

= (\frac{sinθ + 1}{cosθ})^2

 

= (\frac{(sinθ + 1)^2}{cos^2θ}

 

= (\frac{(sinθ + 1) (sinθ + 1)}{1 - sin^2θ }

 

= (\frac{(sinθ + 1) (sinθ + 1)}{(1 - sinθ) (1 + sinθ)}

 

= (\frac{sinθ + 1}{1 - sinθ}

 

= ডানপক্ষ

অতএব, (tanθ + secθ)^2 = \frac{1 + sinθ}{1 - sinθ}

 

১৮। \frac{cotA + cotB}{cotB + tanA} = cotA . tanB

সমাধান: 

বামপক্ষ = \frac{cotA + cotB}{cotB + tanA}

 

= \frac{cotA + cotB}{\frac{1}{tanB} + \frac{1}{cotA}}

 

= \frac{cotA + cotB}{\frac{cotA + tanB}{cotA . tanB}}

 

= (cotA + cotB) \times \frac{cotA . tanB }{cotA + tanB}

= cotA . tanB

 

= ডানপক্ষ

অতএব, \frac{cotA + cotB}{cotB + tanA} = cotA . tanB

 

১৯। \sqrt{\frac{1 - sinA}{1 + sinA}} = secA - tanA

সমাধান: 

বামপক্ষ =  \sqrt{\frac{(1 - sinA) (1 - sinA)}{(1 + sinA)(1 - sinA)}}

 

= \sqrt{\frac{(1 - sinA)^2}{1 - sin^2A}}

 

= \sqrt{\frac{(1 - sinA)^2}{cos^2A}}

 

= \frac{1 - sinA}{cosA}

 

= \frac{1}{cosA} - \frac{sinA}{cosA}

 

= secA – tanA [\because tanA = \frac{sinA}{cosA}, secA = \frac{1}{cosA}]

= ডানপক্ষ

অতএব, \sqrt{\frac{1 - sinA}{1 + sinA}} = secA - tanA

 

২০। \sqrt{\frac{secA + 1}{secA - 1}} = cotA + cosecA

সমাধান: 

বামপক্ষ = \sqrt{\frac{secA + 1}{secA - 1}}

= \sqrt{\frac{(secA + 1)(secA + 1)}{(secA - 1)(secA + 1)}}

= \sqrt{\frac{(secA + 1)^2}{sec^2A - 1}}

= \sqrt{\frac{(secA + 1)^2}{tan^2A}}

= \frac{secA + 1}{tanA}

= \frac{\frac{1}{cosA} + 1}{\frac{sinA}{cosA}}

= \frac{\frac{1 + cosA}{cosA}}{\frac{sinA}{cosA}}

= \frac{\frac{1 + cosA}{cosA}}{\frac{cosA}{sinA}}

= \frac{1 + cosA}{sinA}

= \frac{1}{sinA} + \frac{cosA}{sinA}

= cosecA + cotA [\because \frac{1}{sinA} = cosecA][latex]</span> <span style="font-size: 18pt;">= ডানপক্ষ </span> <span style="font-size: 18pt;">অতএব, [latex]\sqrt{\frac{secA + 1}{secA - 1}} = cotA + cosecA

 

২১। cosA + sinA = \sqrt{2}cosA

সমাধান: দেওয়া আছে, 

(cosA + sinA)^2 = (\sqrt{2}cosA)^2

বা, cos^2A + 2sinA cosA + sin^2A = 2cos^2A

বা, 2cos^2A - cos^2A - 2sinA cosA - sin^2A = 0

বা, cos^2A - 2sinA cosA - sin^2A = 0

বা, cos^2A - 2sinA cosA + sin^2A - 2sin^2A = 0

বা, (cosA - sinA)^2 = 2sin^2A

বা, (cosA - sinA) = \sqrt{2sin^2A}

অতএব, (cosA - sinA) = \sqrt{2} sinA} (প্রমাণিত)

২২। যদি tanA = \frac{1}{\sqrt{3}} হয়, তবে এর মান নির্ণয় কর।

সমাধান: দেওয়া আছে, 

tanA = \frac{1}{\sqrt{3}}

বা, tan^2A = (\frac{1}{\sqrt{3}})^2

বা, tan^2A = \frac{1}{3}

বা, \frac{1}{cot^2A} = \frac{1}{3}

বা, \therefore cot^2A = 3

আমরা জানি, 
cosec^2A = 1 + cot^2A

\therefore cosec^2A = 1 + 3 = 4

sec^2A = 1 + \frac{1}{tan^2A} = 1 + \frac{1}{3} = \frac{4}{3}

প্রদত্ত রাশি, 

\frac{cosec^2A - sec^2A}{cosec^2A + sec^2A}

\frac{4 - \frac{4}{3}} {4 + \frac{4}{3}}

= \frac{\frac{12 - 4}{3}}{\frac{12 + 4}{3}}

= \frac{\frac{8}{3}}{\frac{16}{3}}

= \frac{8}{3} \times \frac{3}{16}

= \frac{1}{2}

 

 

 

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