১ম অংশের সমাধান লিংক
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
(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
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></p> <p><span style="font-size: 18pt;">= ডানপক্ষ </span></p> <p><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}
