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2: Polynomials
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5: Arithmetic Progression
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7: Triangles
8: Circles
9: Constructions
Trigonometry
10: Trignometric Ratios
11: T-Ratios of Some Particular Angles
▶ 12: Trigonometric Ratios of Some Complemantary Angles
13: Trigonometric identities
14: Heights and Distances
Mensuration
15: Perimeter And Area of Plane Figures
16: Area of Circle, Sector and Segment
17: Volumes and Surface Areas of Solids
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18: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
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![R.S. Aggarwal solutions for Mathematics [English] Class 10 chapter 12 - Trigonometric Ratios of Some Complemantary Angles R.S. Aggarwal solutions for Mathematics [English] Class 10 chapter 12 - Trigonometric Ratios of Some Complemantary Angles - Shaalaa.com](/images/mathematics-english-class-10_6:8f062ea57bdf49abb4f6d22550b39d56.jpg)
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Solutions for Chapter 12: Trigonometric Ratios of Some Complemantary Angles
Below listed, you can find solutions for Chapter 12 of CBSE, Karnataka Board R.S. Aggarwal for Mathematics [English] Class 10.
R.S. Aggarwal solutions for Mathematics [English] Class 10 12 Trigonometric Ratios of Some Complemantary Angles Exercises [Pages 312 - 314]
Without using trigonometric tables, evaluate :
`sin 16^circ/cos 74^circ`
Without using trigonometric tables, evaluate :
`sec 11^circ/("cosec" 79^circ)`
Without using trigonometric tables, evaluate :
`tan 27^circ/cot 63^circ`
Without using trigonometric tables, evaluate :
`cos 35^circ/sin 55^circ`
Without using trigonometric tables, evaluate :
`("cosec" 42^circ)/sec 48^circ`
Without using trigonometric tables, evaluate :
`cot 38^circ/tan 52^circ`
Without using trigonometric tables, prove that:
cos 81° − sin 9° = 0
Without using trigonometric tables, prove that:
tan 71° − cot 19° = 0
Without using trigonometric tables, prove that:
cosec 80° − sec 10° = 0
Without using trigonometric tables, prove that:
cosec272° − tan218° = 1
Without using trigonometric tables, prove that:
cos275° + cos215° = 1
Without using trigonometric tables, prove that:
tan266° − cot224° = 0
Without using trigonometric tables, prove that:
sin248° + sin242° = 1
Without using trigonometric tables, prove that:
cos257° − sin233° = 0
Without using trigonometric tables, prove that:
(sin 65° + cos 25°)(sin 65° − cos 25°) = 0
Without using trigonometric tables, prove that:
sin53° cos37° + cos53° sin37° = 1
Without using trigonometric tables, prove that:
cos54° cos36° − sin54° sin36° = 0
Without using trigonometric tables, prove that:
sec70° sin20° + cos20° cosec70° = 2
Without using trigonometric tables, prove that:
sin35° sin55° − cos35° cos55° = 0
Without using trigonometric tables, prove that:
(sin72° + cos18°)(sin72° − cos18°) = 0
Without using trigonometric tables, prove that:
tan48° tan23° tan42° tan67° = 1
Prove that:
`(sin 70^circ)/(cos 20^circ) + ("cosec" 20^circ)/(sec 70^circ) - 2 cos 70^circ "cosec" 20^circ = 0`
Prove that:
`cos 80^circ/(sin 10^circ) + cos 59^circ "cosec" 31^circ = 2`
Prove that:
`(2 "sin" 68^circ)/(cos 10^circ )- (2 cot 15^circ)/(5 tan 75^circ) = ((3 tan 45^circ t an 20^circ tan 40^circ tan 50^circ tan 70^circ)) /5= 1`
Prove that:
`sin 18^circ/(cos 72^circ )+ sqrt(3)(tan 10^circ tan 30^circ tan 40^circ tan50^circ tan 80^circ) `
Prove that:
sin θ cos (90° - θ ) + sin (90° - θ) cos θ = 1
Prove that:
\[\frac{\sin\theta}{\cos(90° - \theta)} + \frac{\cos\theta}{\sin(90° - \theta)} = 2\]
Prove that:
\[\frac{\sin\theta \cos(90^\circ - \theta)\cos\theta}{\sin(90^\circ- \theta)} + \frac{\cos\theta \sin(90^\circ - \theta)\sin\theta}{\cos(90^\circ - \theta)}\]
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
Prove that:
\[\frac{\cos(90^\circ - \theta)}{1 + \sin(90^\circ - \theta)} + \frac{1 + \sin(90^\circ- \theta)}{\cos(90^\circ - \theta)} = 2 cosec\theta\]
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
Prove that:
\[cot\theta \tan\left( 90° - \theta \right) - \sec\left( 90° - \theta \right)cosec\theta + \sqrt{3}\tan12° \tan60° \tan78° = 2\]
Prove that :
tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]
Prove that:
cot12° cot38° cot52° cot60° cot78° = \[\frac{1}{\sqrt{3}}\]
Prove that:
cos15° cos35° cosec55° cos60° cosec75° = \[\frac{1}{2}\]
Prove that:
cos1° cos2° cos3° ... cos180° = 0
Prove that:
\[\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2\]
Prove that
sin (70° + θ) − cos (20° − θ) = 0
Prove that
tan (55° − θ) − cot (35° + θ) = 0
Prove that
cosec (67° + θ) − sec (23° − θ) = 0
Prove that
cosec (65 °+ θ) sec (25° − θ) − tan (55° − θ) + cot (35° + θ) = 0
Prove that
sin (50° + θ ) − cos (40° − θ) + tan 1° tan 10° tan 80° tan 89° = 1.
Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.
sin67° + cos75°
Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.
cot65° + tan49°
Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.
sec78° + cosec56°
Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.
cosec54° + sin72°
If A, B and C are the angles of a ΔABC, prove that tan `((C + "A")/2) = cot B/2`
If cos 20 = sin 4 θ ,where 2 θ and 4 θ are acute angles, then find the value of θ
If sec2A = cosec(A - 42°), where 2A is an acute angle, then find the value of A.
If sin 3 A = cos (A − 26°), where 3 A is an acute angle, find the value of A.
If tan 2 A = cot (A − 12°), where 2 A is an acute angle, find the value of A.
If sec 4 A = cosec (A − 15°), where 4 A is an acute angle, find the value of A.
Solutions for 12: Trigonometric Ratios of Some Complemantary Angles
![R.S. Aggarwal solutions for Mathematics [English] Class 10 chapter 12 - Trigonometric Ratios of Some Complemantary Angles R.S. Aggarwal solutions for Mathematics [English] Class 10 chapter 12 - Trigonometric Ratios of Some Complemantary Angles - Shaalaa.com](/images/mathematics-english-class-10_6:8f062ea57bdf49abb4f6d22550b39d56.jpg)
R.S. Aggarwal solutions for Mathematics [English] Class 10 chapter 12 - Trigonometric Ratios of Some Complemantary Angles
Shaalaa.com has the CBSE, Karnataka Board Mathematics Mathematics [English] Class 10 CBSE, Karnataka Board solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. R.S. Aggarwal solutions for Mathematics Mathematics [English] Class 10 CBSE, Karnataka Board 12 (Trigonometric Ratios of Some Complemantary Angles) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.
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Concepts covered in Mathematics [English] Class 10 chapter 12 Trigonometric Ratios of Some Complemantary Angles are Trigonometric Ratios, Trigonometric Ratios of Specific Angles, Trigonometric Identities (Square Relations), Trigonometry, Relation Among Trigonometric Ratios, Trigonometric Ratios, Trigonometric Ratios of Specific Angles, Trigonometric Identities (Square Relations), Trigonometry, Relation Among Trigonometric Ratios.
Using R.S. Aggarwal Mathematics [English] Class 10 solutions Trigonometric Ratios of Some Complemantary Angles exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in R.S. Aggarwal Solutions are essential questions that can be asked in the final exam. Maximum CBSE, Karnataka Board Mathematics [English] Class 10 students prefer R.S. Aggarwal Textbook Solutions to score more in exams.
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