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 \[[\because \frac{1}{sinA} = cosecA][latex]
= ডানপক্ষ
অতএব, [latex]\sqrt{\frac{secA + 1}{secA – 1}} = cotA + cosecA\]
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,
\[\frac{cosec^2A – sec^2A}{cosec^2A + sec^2A}\]
\[\frac{4 – \frac{4}{3}} {4 + \frac{4}{3}}\]
= \[\frac{\frac{12 – 4}{3}}{\frac{12 + 4}{3}}\]
= \[\frac{\frac{8}{3}}{\frac{16}{3}}\]
= \[\frac{8}{3} \times \frac{3}{16}\]
= \[\frac{1}{2}\]
