SEBA Class 10 Maths Exercise 8.4 ত্ৰিকোণমিতিৰ পৰিচয় Assamese Medium
এই পোষ্টটোত দশম শ্ৰেণীৰ গণিত পাঠ্যপুথিৰ Chapter 8 ত্ৰিকোণমিতিৰ পৰিচয় পাঠটোৰ অনুশীলনী 8.4 ৰ প্ৰশ্নবোৰৰ সমাধান লাভ কৰিব। অন্য অনুশীলনীবোৰৰ সমাধান পাবলৈ তলত দিয়া লিংকত ক্লিক কৰিব। ত্ৰিকোণমিতিৰ পৰিচয় পাঠটিত মুঠতে চাৰিটা অনুশীলনী আছে - Exercise 8.1, Exercise 8.2, Exercise 8.3, Exercise 8.4
প্ৰতিটো পাঠৰ MCQs আৰু সমাধান পাবলৈ ইয়াত ক্লিক কৰিব - MCQs.
দশম শ্ৰেণীৰ গণিতৰ পাথ্যপুথিত থকা পাঠসমূহ হ'ল পুনৰালোচনা (Revision), বাস্তৱ সংখ্যা (Real Numbers), বহুপদ (Polynomials), দুটা চলকত ৰৈখিক সমীকৰণৰ যোৰ (Pair of Linear Equations in Two Variables), দ্বিঘাত সমীকৰণ (Quadratic Equations), সমান্তৰ প্ৰগতি (Arithmetic Progressions), ত্ৰিভুজ (Triangles), স্থানাংক জ্যামিতি (Coordinate Geometry), ত্ৰিকোণমিতিৰ পৰিচয় (Introduction to Trigonometry), ত্ৰিকোণমিতিৰ কিছুমান প্ৰয়োগ (Some Applications of Trigonometry), বৃত্ত (Circles), অংকন (Constructions), বৃত্ত সম্বন্ধীয় কালি (Areas Related to Circles), পৃষ্ঠকালি আৰু আয়তন (Surface Areas and Volumes), পৰিসংখ্যা (Statistics), সম্ভাৱিতা (Probability).
Exercise Solutions of Chapter 8 Introduction to Trigonometry
- Class 10 Chapter 8 Exercise 8.1 Solutions
- Class 10 Chapter 8 Exercise 8.2 Solutions
- Class 10 Chapter 8 Exercise 8.3 Solutions
Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.4 in Assamese Medium
1.sinA, secA আৰু tanA এই ত্ৰিকোণমিতিক অনুপাত কেইটাক cotA ৰ দ্বাৰা প্ৰকাশ কৰা।
Solution:
(i) আমি জানো যে,
$\LARGE cosec^2 A = 1+cot^2 A$
$ \LARGE ⇒ cosecA = \sqrt{1 + cot^2A}$
$\LARGE ⇒ sinA = \frac{1}{\sqrt {1 + cot^2 A}}$
আকৌ,
$\LARGE sec^2A = 1 + tan^2A$
$\LARGE ⇒ sec^2A = 1+\frac {1}{cot^2A}$
$\LARGE ⇒ secA = \sqrt{1+\frac {1}{cot^2A}}$
$\LARGE ⇒ secA = \sqrt{\frac {cot^2A + 1}{cot^2A}}$
$\LARGE ⇒ secA = \frac{\sqrt{cot^2A + 1}}{cotA}$
আকৌ,
$\LARGE tan A = \frac {1}{cot A}$
2.secA ৰ সহায়ত ∠A কোনৰ আন সকলোবিলাক ত্ৰিকোণমিতিক অনুপাত লিখা।
Solution:
(i) আমি জানো,
$\LARGE cosA =\frac {1}{secA}$ $ \quad \quad$-----(i)
আনহাতে,
$\LARGE sin^2A + cos^2 A =1$
$\LARGE ⇒sin^2A = 1 - cos^2 A $
$\LARGE ⇒sinA = \sqrt{1-cos^2A}$
$\LARGE ⇒sinA = \sqrt {1-(\frac {1}{secA})^2}$ [(i)ৰ পৰা]
$\LARGE ⇒sinA = \sqrt{1-\frac{1}{sec^2A}}$
$\LARGE ⇒sinA = \sqrt {\frac{sec^2A - 1}{sec^2A}}$
$\LARGE ⇒sinA = \frac{\sqrt{sec^2A-1}}{secA}$ $\quad$-----(ii)
আকৌ,
$\LARGE cosecA=\frac {1}{sinA}$
$\LARGE ⇒cosecA = \frac {1}{\frac {\sqrt {sec^2A-1}}{secA}}$ [(ii ৰ পৰা)]
$\LARGE ⇒ cosecA=\frac {secA}{\sqrt {sec^2A - 1}}$
আনহাতে,
$\large 1+tan^2A = sec^2A$
$\large ⇒tan^2A=sec^2A-1$
$\large ⇒tanA = \sqrt{sec^2A-1}$ $\quad$-----(iii)
আকৌ,
$\large cotA=\frac {1}{tanA}$
$\large ⇒cotA = \frac {1}{\sqrt{sec^2-1}}$ $\quad$ [(iii) ৰ পৰা]
3.মান নিৰ্ণয় কৰা-
(i)sin² 63°+sin² 27°/cos² 17°+cos² 73°
(ii)sin 25° cos 65° + cos 25° sin 65°
Solution:
(i)
$\LARGE \frac {sin^2 63° + sin^2 27°}{cos^2 17° + cos^2 73°}$
$\LARGE =\frac {[cos(90°-63°)]^2 + sin^2 27°}{[sin(90°-17°)^2 + cos^2 73°}$
$\LARGE =\frac {cos^2 27° + sin^2 27°}{sin^2 73° + cos^2 73°}$
$\quad \quad$ [$\large \because sin(90°-θ)=cosθ$ & $\large cos(90°-θ) = sinθ$]
$\LARGE =\frac{1}{1}$ $\quad \quad$ [$\because$$ \large sin^2θ+cos^2θ=1$]
$\large =1$
(ii)
$\large sin25°cos65° + cos25° sin 65°$
$\large =sin25° . cos(90°-25°)$ + $\large cos25° . sin(90° - 25°)$
$\large sin25° . sin25° + cos25° . cos25°$
$\quad \quad$ [$\large \because sin(90°-θ)=cosθ$ &$\large cos(90°-θ)=sinθ$]
$\large =sin^2 25° + cos^2 25°$
$\large = 1$
4.শুদ্ধ উত্তৰটো বাছি উলিওৱা। তোমাৰ বাছনিৰ যথাৰ্থতা সাব্যস্ত কৰা।
(i) 9 sec² A - 9 tan²A =
(A)1 $\quad$ (B)9 $\quad$ (C)8 $\quad$ (D)0
(ii) (1+tan θ + sec θ) (1+cot θ - cosec θ) =
(A)0 $\quad$ (B)1 $\quad$ (C)2 $\quad$ (D)-1
(iii) (sec A + tan A) (1 - sin A) =
(A)secA $\quad$ (B)sinA $\quad$ (C)cosecA $\quad$ (D)cosA
(iv) 1+tan²A/1+cot²A=
(A)sec²A $\quad$ (B)-1 $\quad$ (C)cot²A $\quad$ (D)tan²A
Solution:
(i) $\large 9sec^2A - 9tan^2A$
$\large =9 (sec^2A-tan^2A)$
$\large =9×1 \quad \quad$ [$\because$$1+tan^2A=sec^2A$]
$\large = 9$
Correct Option:(B) 9
(ii)
$(1+tanθ+secθ)(1+cotθ-cosecθ)$
$= \large (1 + \frac {sin θ}{cos θ} + \frac{1}{cosθ})$ $\large (1+\frac{cosθ}{sinθ}-\frac{1}{sinθ})$
$= \large (\frac{cosθ+sinθ+1}{cosθ})$ $\large (\frac{sinθ+cosθ-1}{sinθ})$
$\large =\frac{(cosθ+sinθ+1)(sinθ+cosθ-1)}{sinθ. cosθ}$
$\LARGE =\frac{(cosθ+sinθ)^2-1^2}{sinθ. cosθ}$
$\LARGE =\frac{cos^2θ + sin^2θ + 2. cosθ. sinθ - 1}{sin θ. cosθ}$
$\LARGE =\frac {1 + 2 cosθ. sinθ - 1}{sinθ. cosθ}$
$\LARGE =\frac{2. cosθ. sinθ}{sinθ . cosθ}$
$\large = 2$
Correct Option: (C) 2
(iii)
$\large (secA + tanA)(1 - sinA)$
$\LARGE =(\frac {1}{cosA} + \frac {sinA}{cosA}) $$\large (1 - sinA)$
$\LARGE =\frac {(1 + sinA)(1 - sinA)}{cosA}$
$\LARGE =\frac{1^2 - sin^2A}{cosA}$
$\LARGE =\frac {1 - sin^2A}{cosA}$
$\LARGE =\frac {cos^2A}{cosA}$
$\large = cosA$
Correct Option: (D) cosA
(iv)
$\LARGE =\frac {1+tan^2A}{1+cot^2A}$
$\LARGE =\frac {sec^2A}{cosec^2A}$
$\LARGE =\frac {\frac {1}{cos^2A}}{\frac{1}{sin^2A}}$
$\LARGE =\frac{1}{cos^2A} ×$ $\LARGE \frac{sin^2A}{1}$
$\LARGE =\frac {sin^2A}{cos^2A}$
$\large =tan^2A$
Correct Option : (D) $tan^2A$
5. তলৰ অভেদ কেইটা প্ৰমাণ কৰা যদিহে ইয়াত কোণ বিলাক সুক্ষ্ম কোণ আৰু যাৰ বাবে অভেদ কেইটা সংজ্ঞাবদ্ধ হয়-
(i) (cosecθ - cot θ)² = 1 - cosθ / 1 + cosθ
(ii) $\large \frac {cosA}{1 + sinA} + \frac{1+sinA}{cosA} =$ $\large 2secA$
(iii) $\large \frac{tanθ}{1-cotθ} + \frac{cotθ}{1-tanθ} =$ $\large 1+ secθ.cosecθ$
[ইংগিত : ইয়াত থকা পদবোৰ sinθ আৰু cosθ ত প্ৰকাশ কৰা]
(iv) $\large \frac{1+secA}{secA} =$$\large \frac{sin^2A}{1-cosA}$ [ইংগিত: বাওঁপক্ষ আৰু সোঁপক্ষ বেলেগে সৰল কৰা।]
(v) $\large \frac{cosA - sin A + 1}{cos A + sin A - 1} =$$\large cosec A + cot A$, $\large cosec² A = 1 + cot² A$ অভেদৰ সহায়ত কৰা।
(vi) $\large \sqrt{\frac{1+sinA}{1-sinA}} =$$\large sec A + tan A$
(vii) $\large \frac{sinθ - 2sin^3θ}{2 cos^3θ - cosθ} =$$\large tanθ$
(viii) $(sinA + cosecA)^2 + (cosecA + secA)^2 =$$ 7 + tan^2A + cot^2A$
(ix) $(cosec A - sinA)(sec A - cos A)=$$\large \frac{1}{tanA + cotA}$
[ইংগিত: বাওঁপক্ষ আৰু সোঁপক্ষ বেলেগে সৰল কৰা]
(x) $\large \left( \frac{1+tan^2A}{1+cot^2A} \right)=$ $\large \left( \frac{1-tanA}{1-cotA} \right)^2 = $$ tan^2A$
Solution:
(i)
$\large (cosecθ - cotθ)^2 = \frac {1-cosθ}{1+cosθ}$
L.H.S. $\large = (cosecθ - cotθ)^2$
$\LARGE =(\frac{1}{sinθ}-\frac{cosθ}{sinθ})^2$
$\LARGE =(\frac{1}{sinθ}-\frac{cosθ}{sinθ})^2$
$\LARGE =(\frac{1-cosθ}{sinθ})^2$
$\LARGE =\frac{(1-cosθ)^2}{sin^2θ}$
$\LARGE =\frac{(1-cosθ)^2}{1-cos^2θ}$
$\LARGE =\frac{(1-cosθ)(1-cosθ)}{(1+cosθ)(1-cosθ)}$
$\LARGE =\frac {1-cosθ}{1+cosθ}$
= R.H.S.
(ii) $\LARGE \frac {cosA}{1+sinA}+\frac{1+sinA}{cosA} = $$\large 2secA$
L.H.S $\LARGE = \frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}$
$\LARGE =\frac {cos^2 + (1+sinA)^2}{(1+sinA). cosA}$
$\LARGE =\frac{cos^2A + 1 + sin^2A + 2. sinA}{(1+sinA). cosA}$
$\LARGE =\frac {1+1+2. sinA}{(1+sinA). cosA}$
$\LARGE =\frac {2+2. sinA}{(1+sinA). cosA}$
$\LARGE =\frac {2(1+sinA)}{(1+sinA). cosA}$
$\LARGE =\frac {2}{cosA}$
$\LARGE =2\times\frac{1}{cosA}$
$\large = 2.secA$
= R.H.S.
(iii)
$\LARGE \frac{tanθ}{1-cotθ}+\frac{cotθ}{1-tanθ}=$$\large 1+secθ. cosecθ$
L.H.S $\LARGE = \frac {tanθ}{1-cotθ}+\frac{cotθ}{1-tanθ}$
$\LARGE = \frac{\frac{sinθ}{cosθ}}{1-\frac{cosθ}{sinθ}} + \frac{ \frac{cosθ}{sinθ}}{1-\frac{sinθ}{cosθ}}$
$\LARGE =\frac{\frac{sinθ}{cosθ}}{\frac{sinθ-cosθ}{sinθ}} + \frac{\frac{cosθ}{sinθ}}{\frac{cosθ-sinθ}{cosθ}}$
$\LARGE =\frac {sinθ}{cosθ} \times \frac{sinθ}{sinθ-cosθ} +$$\LARGE \frac {cosθ}{sinθ} \times \frac {cosθ}{cosθ-sinθ}$
$\LARGE =\frac{sin^2θ}{cosθ(sinθ-cosθ)} - $$\LARGE \frac{cos^2θ}{sinθ(sinθ-cosθ)}$
$\LARGE =\frac{1}{sinθ-cosθ}$$\LARGE \left( \frac {sin^2θ}{cosθ}-\frac{cos^2θ}{sinθ} \right)$
$\LARGE =\frac {1}{sinθ. cosθ}$$\LARGE \left( \frac{sin^3θ - cos^3θ}{sinθ. cosθ} \right)$
$\large =\frac{1}{sinθ-cosθ}$ $\times$ $\large \frac{(sinθ-cosθ)(sin^2θ+cos^2θ+sinθ. cosθ)}{sinθ. cosθ}$ $\quad \quad$ $\large [∵ a^3-b^3=(a-b)(a^2+b^2+ab)]$
$\LARGE =\frac{1+sinθ. cosθ}{sinθ. cosθ}$
$\LARGE =\frac{1}{sinθ. cosθ}+\frac{sinθ. cosθ}{sinθ. cosθ}$
$\large =1+secθ. cosecθ$
= R.H.S.
(iv)
$\LARGE \frac{1+secA}{secA}=\frac{sin^2 A}{1-cosA}$
L.H.S. = $\large \frac{1+secA}{secA}$
= $\large \frac{1}{secA} + \frac{secA}{secA}$
= cosA + 1
R.H.S. = $\frac{sin^2 A}{1-cosA}$
= $\frac{1 - cos^2 A}{1-cosA}$
= $\frac{(1 - cosA)(1 + cosA)}{1-cosA}$
= cosA + 1
∴ L.H.S. = R.H.S.
বিকল্প পদ্ধতি
L.H.S $\LARGE = \frac {1+secA}{secA}$
$\LARGE =\frac {1}{secA}+\frac{secA}{secA}$
$\large = cosA+1$
$\LARGE =\frac {(1+cosA)(1-cosA)}{1 (1-cosA)}$
$\qquad \qquad $[লৱ আৰু হৰক (1-cosA) ৰে পূৰণ কৰি]
$\LARGE =\frac {1-cos^2A}{1-cosA}$
$\LARGE =\frac {sin^2A}{1-cosA}$
= R.H.S.
(v) $\LARGE \frac {cosA-sinA + 1}{cosA+sinA-1} = $$\large cosecA + cotA$
L.H.S. $\LARGE = \frac {cosA-sinA+1}{cosA + sinA - 1}$
$\large =\frac {\frac{cosA}{sinA}-\frac{sinA}{sinA}+\frac{1}{sinA}}{\frac{cosA}{sinA}+\frac{sinA}{sinA}-\frac{1}{sinA}}$
$\quad \quad$ [লৱ আৰু হৰক sinA ৰে হৰণ কৰি ]
$\large =\frac{cotA-1+cosecA}{cotA+1-cosecA}$
$\large =\frac {cotA-cosec^2+cot^2+cosecA}{cotA+1-cosecA}$ $\quad \quad$ $[\large \because cosec^2A - cot^2A = 1]$
$\large =\frac{cotA+cosecA - (cosec^2A - cot^2A)}{cotA+1-cosecA}$
$=\frac{(cotA+cosecA)-(cosecA+cotA)(cosecA-cotA)}{cotA+1-cosecA}$
$\large =\frac{(cotA+cosecA)(1-cosec+cotA)}{(cotA+1-cosecA)}$
$\large =cotA + cosecA$
= R.H.S.
(vi) $\LARGE \sqrt{\frac{1+sinA}{1-sinA}} = $$\large secA + tanA$
L.H.S. $\LARGE = \sqrt{\frac{1+sinA}{1-sinA}}$
$\LARGE =\sqrt{\frac{(1+sinA)(1+sinA)}{(1-sinA)(1+sinA)}}$
$\quad \quad$ [ লৱ আৰু হৰক (1+sinA) ৰে পূৰণ কৰি]
$\LARGE =\sqrt{\frac{(1+sinA)^2}{1-sin^2A}}$
$\LARGE =\frac{1+sinA}{\sqrt{cos^2A}}$
$\LARGE =\frac {1+sinA}{cosA}$
$\LARGE =\frac {1}{cosA}+\frac{sinA}{cosA}$
$\large = secA + tanA$
= R.H.S.
(vii) $\LARGE \frac {sinθ-2sin^3θ}{2cos^3-cosθ} =$$\large tanθ$
L.H.S = $\LARGE \frac {sinθ - 2sin^3θ}{2cos^3θ - cosθ}$
$\LARGE =\frac{sinθ(1-2sin^2θ)}{cosθ(2cos^2θ -1)}$
$\LARGE =\frac {sinθ(1-2sin^2θ)}{cosθ[2(1-sin^2θ)-1]}$
$\LARGE =\frac{sinθ(1-2sin^2θ)}{cosθ (2-2sin^2θ)-1}$
$\LARGE =\frac{sinθ(1-2sin^2θ)}{cosθ(1-2sin^2θ)}$
$\LARGE =\frac{sinθ}{cosθ}$
$\large = tanθ$
=R.H.S
(viii)
$(sinA+cosecA)^2+(cosA+secA)^2 = $ $7+tan^2A+cot^2A$
L.H.S. $ = (sinA+cosecA)^2 + (cosA+secA)^2$
$ =sin^2A + cosec^2A+2.sinA.cosecA + $ $ cos^2A+sec^2A+2.cosA.secA$
$ =sin^2A+cos^2A + 1 +$ $cot^2A+2.sinA. $$\large \frac{1}{sinA}$ $+1+tan^2A +$ $2.cosA. $$\large \frac{1}{cosA}$
$=1+1+cot^2A+2+1+$$tan^2A+2$
$=7+tan^2A+cot^2A$
= R.H.S.
(ix) $(cosecA-sinA)(secA-cosA)$ = $\large \frac{1}{tanA+cotA}$
L.H.S. = $\large (cosecA-sinA)(secA-cosA)$
= $\large (\frac{1}{sinA}-sinA)(\frac{1}{cosA}-cosA)$
= $\large (\frac{1 - sin^2 A}{sinA})(\frac{1 - cos^2 A}{cosA})$
= $\large \frac{cos^2 A}{sinA} × \frac{sin^2 A}{cosA}$
= cosA. sinA
R.H.S. = $\large \frac{1}{tanA+cotA}$
= $\LARGE \frac{1}{\frac{sinA}{cosA} + \frac{cosA}{sinA}}$
= $\LARGE \frac{1}{\frac{sin^2A + cos^2A}{cosA.SinA}}$
= $\large \frac{cosA.sinA}{1}$
= cosA.sinA
∴ L.H.S. = R.H.S.
বিকল্প পদ্ধতি
L.H.S. $ = (cosecA - sinA)(secA-cosA)$
$\large =(\frac{1}{sinA}$$-sinA).$$\large (\frac{1}{cosA}$$-cosA)$
$\LARGE =(\frac{1-sin^2A}{sinA})(\frac{1-cos^2A}{cosA})$
$\LARGE =\frac {cos^2A}{sinA}$ $\times$ $\LARGE \frac{sin^2A}{cosA}$
$\LARGE =\frac{sinA. cosA}{1}$
$\LARGE =\frac{sinA. cosA}{sin^2A+cos^2A}$
$\LARGE =\frac {1}{\frac{sin^2A+cos^2A}{sinA. cosA}}$
$\quad \quad$ [ লৱ আৰু হৰক sinA. cosA ৰে হৰণ কৰিলে ]
$\LARGE =\frac{1}{\frac{sin^2A}{sinA. cosA}+\frac{cosA^2A}{sinA. cosA}}$
$\LARGE =\frac{1}{\frac{sinA}{cosA}+\frac{cosA}{sinA}}$
$\LARGE =\frac{1}{tanA+cotA}$
= R.H.S.
(x) $\large \left( \frac{1+tan^2A}{1+cot^2A} \right)=$ $\large \left( \frac{1-tanA}{1-cotA} \right)^2 = $$ tan^2A$
L.H.S. $\LARGE = \frac{1+tan^2A}{1+cot^2A}$
$\LARGE =\frac{sec^2A}{cosec^2A}$
$\LARGE =\frac {\frac{1}{cos^2A}}{\frac{1}{sin^2A}}$
$\LARGE =\frac{1}{cos^2A}$ $\times$ $\LARGE \frac{sin^2A}{1}$
$\large =tan^2A$
= R.H.S.
একেদৰে,
L.H.S. $\LARGE = \left( \frac{1-tanA}{1-cotA} \right)^2$
$\LARGE = \left( \frac{1-tanA.}{1-\frac{1}{tanA}} \right)^2$
$\LARGE = \left( \frac{1-tanA}{\frac{tanA-1}{tanA}} \right)^2$
$\large = (1-tanA)^2 ×$$\LARGE \frac{tan^2A}{(1-tanA)^2} $
$\large =tan^2A$
= R.H.S.
6. প্ৰমাণ কৰা:
(i) $\large tan^4θ + tan^2θ = sec^4θ - sec^2θ$
(ii) $\LARGE \frac{cosθ}{1-tanθ} + \frac{sinθ}{1-cotθ} =$$\large sinθ + cosθ$
(iii) $\LARGE \sqrt{\frac {secθ-1}{secθ+1}} =$$\large cosecθ-cotθ$
(iv) $\large cotθ + tanθ = secθ.cosecθ$
(v) $\LARGE \frac{1}{1+sinθ} + \frac{1}{1-sinθ} = $$\large 2sec^2θ$
Solution:
(i)
$\large tan^4θ + tan^2θ = sec^4θ - sec^2θ$
L.H.S. $\large = tan^4θ + tan^2θ$
$\large =tan^2θ(tan^2θ+1)$
$\large =tan^2θ . sec^2θ$
$\large =(sec^2θ-1) . sec^2θ$
$\large =sec^4θ - sec^2θ$
= R.H.S.
(ii)
$\LARGE \frac{cosθ}{1-tanθ} + \frac{sinθ}{1-cotθ} =$$\large sinθ + cosθ$
L.H.S $\LARGE = \frac{cosθ}{1-tanθ}+\frac{sinθ}{1-cotθ}$
$\LARGE = \frac{cosθ}{1- \frac{sinθ}{cosθ}}+\frac{sinθ}{1- \frac{cosθ}{sinθ}}$
$\LARGE =\frac{cosθ}{\frac{cosθ-sinθ}{cosθ}}+\frac{sinθ}{\frac{sinθ-cosθ}{sinθ}}$
$\LARGE =\frac{cos^2θ}{cosθ-sinθ}-\frac{sin^2θ}{cosθ-sinθ}$
$\LARGE =\frac{cos^2θ-sin^2θ}{cosθ-sinθ}$
$\LARGE =\frac{(cosθ-sinθ)(cosθ+sinθ)}{(cosθ-sinθ)}$
$\large = sinθ + cosθ$
= R.H.S.
(iii)
$\LARGE \sqrt{\frac {secθ-1}{secθ+1}} =$$\large cosecθ-cotθ$
L.H.S $\LARGE = \sqrt{\frac{secθ -1}{secθ+1}}$
$\LARGE =\sqrt{\frac{(secθ-1)(secθ-1)}{(secθ+1)(secθ-1)}}$
$\LARGE =\sqrt{\frac{(secθ-1)^2}{sec^2θ-1}}$
$\LARGE =\frac{secθ-1}{\sqrt{tan^2θ}}$
$\LARGE =\frac{secθ-1}{tanθ}$
$\LARGE =\frac{secθ}{tanθ}-\frac{1}{tanθ}$
$\LARGE =\frac{1}{cosθ}$ $\times$ $\LARGE \frac{cosθ}{sinθ} - $$\large cotθ$
$\LARGE =\frac{1}{sinθ} - $$\large cotθ$
$\large = cosecθ-cotθ$
= R.H.S.
(iv) $\large cotθ + tanθ = secθ.cosecθ$
L.H.S. $\large = cotθ + tanθ$
$\LARGE =\frac{cosθ}{sinθ}+\frac{sinθ}{cosθ}$
$\LARGE =\frac{cos^2θ + sin^2θ}{sinθ.cosθ}$
$\LARGE =\frac{1}{sinθ.cosθ}$
$\LARGE =\frac{1}{sinθ}$ $\times$ $\LARGE \frac{1}{cosθ}$
$\large = cosecθ. secθ$
= R.H.S.
(v)
$\LARGE \frac{1}{1+sinθ} + \frac{1}{1-sinθ} = $$\large 2sec^2θ$
L.H.S. $\LARGE = \frac{1}{1+sinθ}+\frac{1}{1-sinθ}$
$\LARGE =\frac{1-sinθ+1+sinθ}{(1+sinθ)(1-sinθ)}$
$\LARGE =\frac{2}{1-sin^2θ}$
$\LARGE =\frac{2}{cos^2θ}$
$\large =2sec^2θ$