Introduction to Trigonometry (ತ್ರಿಕೋನಮಿತಿಯ ಪ್ರಸ್ತಾವನೆ) - Formulas and Model Questions:
Formulas, Model Questions, Solved Questions from the previous years' questions papers on Introduction to Trigonometry (ತ್ರಿಕೋನಮಿತಿಯ ಪ್ರಸ್ತಾವನೆ) for the SSLC Kannada medium students
have been updated in this post below. The students of SSLC can make use of this Online Study Package to get good scores in the SSLC examinations.Teachers also can help the students to access this platform to use this Online Study Package anywhere and any time.
ತ್ರಿಕೋನಮಿತಿಯ ಪ್ರಸ್ತಾವನೆ
ಮುಖ್ಯಾಂಶಗಳು
ಕೋನಗಳ ಅನುಪಾತ
| sin θ |
ಅಭಿಮುಖ ಬಾಹು
/
ವಿಕರ್ಣ
|
BC
/
AC
|
| cos θ |
ಪಾರ್ಶ್ವ ಬಾಹು
/
ವಿಕರ್ಣ
|
AB
/
AC
|
| tan θ |
ಅಭಿಮುಖ ಬಾಹು
/
ಪಾರ್ಶ್ವ ಬಾಹು
|
BC
/
AB
|
| cosec θ |
ವಿಕರ್ಣ
/
ಅಭಿಮುಖ ಬಾಹು
|
AC
/
BC
|
| sec θ |
ವಿಕರ್ಣ
/
ಪಾರ್ಶ್ವ ಬಾಹು
|
AC
/
AB
|
| cot θ |
ಪಾರ್ಶ್ವ ಬಾಹು
/
ಅಭಿಮುಖ ಬಾಹು
|
AB
/
BC
|
ವಿಲೋಮ ಅನುಪಾತಗಳು
| sin θ |
1
/
cosec θ
|
| cos θ |
1
/
sec θ
|
| tan θ |
1
/
cot θ
|
| cosec θ |
1
/
sin θ
|
| sec θ |
1
/
cos θ
|
| cot θ |
1
/
tan θ
|
ನಿರ್ದಿಷ್ಟ ಕೋನಗಳಿಗೆ ತ್ರಿಕೋನಮಿತಿಯ ಅನುಪಾತಗಳು
| ಕೋನಗಳು | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin θ | 0 |
1
/
2
|
1
/
√
2
|
√
3
/
2
|
1 |
| cos θ | 1 |
√
3
/
2
|
1
/
√
2
|
1
/
2
|
0 |
| tan θ | 0 |
1
/
√
3
|
1 | √ 3 | ND |
| cosec θ | ND | 2 | √ 2 |
2
/
√
3
|
1 |
| sec θ | 1 |
2
/
√
3
|
√ 2 | 2 | ND |
| cot θ | ND | √ 3 | 1 |
1
/
√
3
|
0 |
ತ್ರಿಕೋನಮಿತಿಯ ಪೂರಕ ಕೋನಗಳು
| sin (90-θ) | cos θ |
| cos (90-θ) | sin θ |
| tan (90-θ) | cot θ |
| cot (90-θ) | tan θ |
| sec (90-θ) | cosec θ |
| cosec (90-θ) | sec θ |
ತ್ರಿಕೋನಮಿತಿಯ ನಿತ್ಯ ಸಮೀಕರಣಗಳು
| sin² θ + cos² θ = 1 |
| tan² θ + 1 = sec² θ |
| 1 + cot² θ = cosec² θ |
ಮಾದರಿ ಪ್ರಶ್ನೆಗಳು
ಎರಡು ಅಂಕದ ಪ್ರಶ್ನೆಗಳು
ಉತ್ತರ:
sin² A + cos² A = 1
0 + cos² A = 1
cos² A = 1
cos A = √1
cos A = 1
sin² A + cos² A = 1
0 + cos² A = 1
cos² A = 1
cos A = √1
cos A = 1
ಉತ್ತರ:
sec θ =
=
=
sec θ =
1
/
Cos θ
=
1
/
24⁄25
=
25
/
24
ಉತ್ತರ:
sec² 26° – tan² 26°
tan² A + 1 = sec² A
= tan² 26° + 1 – tan² 26°
= 1
sec² 26° – tan² 26°
tan² A + 1 = sec² A
= tan² 26° + 1 – tan² 26°
= 1
ಉತ್ತರ:
tan 45° + cot 45° = 1 + 1
tan 45° + cot 45° = 2
tan 45° + cot 45° = 1 + 1
tan 45° + cot 45° = 2
ಉತ್ತರ:
sin A . cos A . tan A + cos A . sin A . cot A
= sin A . cos A .
{ tan A =
= sin A . sin A + cos A . cos A {sin² A + cos² A = 1}
= sin² A + cos² A
= 1
sin A . cos A . tan A + cos A . sin A . cot A
= sin A . cos A .
sin A
/
cos A
+ cos A . sin A .
cos A
/
sin A
{ tan A =
sin A
/
cos A
, cot A =
cos A
/
sin A
}= sin A . sin A + cos A . cos A {sin² A + cos² A = 1}
= sin² A + cos² A
= 1
ಉತ್ತರ:
(1 + tan² 60°)² = (1 + (√3)²)² {tan 60° = √3 }
= (1 + 3)²
= 4²
= 16
(1 + tan² 60°)² = (1 + (√3)²)² {tan 60° = √3 }
= (1 + 3)²
= 4²
= 16
ಉತ್ತರ:
Sin θ =
Cosec θ =
3 cosec θ = 3 *
3 cosec θ = 5
Sin θ =
3
/
5
Cosec θ =
5
/
3
{cosec θ =
1
/
Sin θ
}3 cosec θ = 3 *
5
/
3
3 cosec θ = 5
ಉತ್ತರ:
2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + (√3⁄2)² - (√3⁄2)²
= 2 +
= 2
2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + (√3⁄2)² - (√3⁄2)²
= 2 +
3
/
4
-
3
/
4
= 2
ಉತ್ತರ:
tan 48° * tan 23° * tan 42° * tan 67°
= tan 48° * cot (90°- 42°) * tan 67°* cot (90°- 23°) { cot (90°- θ) = tanθ}
= tan 48° * cot 48° * tan 67° * cot 67°
= tan 48° *
= 1
tan 48° * tan 23° * tan 42° * tan 67°
= tan 48° * cot (90°- 42°) * tan 67°* cot (90°- 23°) { cot (90°- θ) = tanθ}
= tan 48° * cot 48° * tan 67° * cot 67°
= tan 48° *
1
/
tan 48°
* tan 67° *
1
/
tan 67°
= 1 * 1 {cot θ =
1
/
tan θ
}= 1
ಉತ್ತರ :
sin 30° = cos 60° =
tan 45° = 1
cosec 60° = sec 30° =
=
sin 30° = cos 60° =
1
/
2
tan 45° = 1
cosec 60° = sec 30° =
2
/
√3
(sin 30 + tan 45 - cosec 60)
/
(sec 30 + cos 60 + cot 45)
=
1
/
2
+ 1 -
2
/
√3
/
2
/
√3
+
1
/
2
+ 1
=
3√3-2
/
4+3√3
ಉತ್ತರ :
1 - sin² A = cos² A
1 - tan² A = sec² A
(1 - sin² A)(1 - tan² A) = cos² A * sec² A
= cos² A *
= 1
1 - sin² A = cos² A
1 - tan² A = sec² A
(1 - sin² A)(1 - tan² A) = cos² A * sec² A
= cos² A *
1
/
cos² A
{sec² A =
1
/
cos² A
} = 1
ಉತ್ತರ :
Sin 30° =
cos 60° =
tan 45° = 1
Sin 30° x cos 60° – tan² 45° =
=
Sin 30° =
1
/
2
cos 60° =
1
/
2
tan 45° = 1
Sin 30° x cos 60° – tan² 45° =
1
/
2
x
1
/
2
- 1²
1
/
4
- 1=
-3
/
4
ಉತ್ತರ :
=
=
=
= sec² θ + tan² θ + 2 *
= sec² θ + tan² θ + 2 * tan θ * sec θ
= (sec θ + tan θ)²
(1+sin θ)
/
(1-sin θ)
=
(1+sin θ)
/
(1-sin θ)
*
(1+sin θ)
/
(1+sinθ)
=
(1+sin θ)²
/
(1-sin² θ)
=
(1+2sin θ+sin² θ)
/
cos² θ)
=
1
/
cos² θ
+
sin² θ
/
cos² θ
+
2sin θ
/
cos² θ
= sec² θ + tan² θ + 2 *
1
/
cos θ
*
sin θ
/
cos θ)
= sec² θ + tan² θ + 2 * tan θ * sec θ
= (sec θ + tan θ)²
ಉತ್ತರ:
√3 tan θ = 1
tan θ =
tan θ = tan 30° θ = 30°
sin 3θ + cos 2θ = sin (3*30) + cos (2*30)
= sin 90° + cos 60°
= 1 +
= 3/2
√3 tan θ = 1
tan θ =
1
/
√3
tan θ = tan 30° θ = 30°
sin 3θ + cos 2θ = sin (3*30) + cos (2*30)
= sin 90° + cos 60°
= 1 +
1
/
2
{sin 90° = 1, cos60° =
1
/
2
}= 3/2
ಉತ್ತರ:
(tan A * sin A ) + cos A =
=
{sin² A + cos² A = 1}
=
= sec A
(tan A * sin A ) + cos A =
sin A
/
cos A
* sin A + cos A=
sin² A
/
cos A
+ cos A
=
sin² A + cos² A
/
cos A
{sin² A + cos² A = 1}
=
1
/
cos A
= sec A
ಉತ್ತರ:
∆ABC ಯಲ್ಲಿ ∠B = 90°
ಫೈಥಾಗೊರಸ್ ಪ್ರಮೇಯದ ಪ್ರಕಾರ,
AC² = AB² + BC²
4² = AB² + 3²
16 = AB² + 9
16 - 9 = AB²
7 = AB²
AB = √7
COS A =
COS A =
∆ABC ಯಲ್ಲಿ ∠B = 90°
ಫೈಥಾಗೊರಸ್ ಪ್ರಮೇಯದ ಪ್ರಕಾರ,
AC² = AB² + BC²
4² = AB² + 3²
16 = AB² + 9
16 - 9 = AB²
7 = AB²
AB = √7
COS A =
AB
/
BC
COS A =
√3
/
4
ಉತ್ತರ:
∆ABC ಯಲ್ಲಿ ∠B = 90°
ಆದ್ದರಿಂದ ಪೈಥಾಗೊರಸ್ ಪ್ರಮೇಯ ಪ್ರಕಾರ,
AC² = AB² + BC²
169 = 25 + BC²
169 – 25 = BC²
144 = BC² , BC² = √144 , BC = 12
cos θ =
tan θ =
∆ABC ಯಲ್ಲಿ ∠B = 90°
ಆದ್ದರಿಂದ ಪೈಥಾಗೊರಸ್ ಪ್ರಮೇಯ ಪ್ರಕಾರ,
AC² = AB² + BC²
169 = 25 + BC²
169 – 25 = BC²
144 = BC² , BC² = √144 , BC = 12
cos θ =
ಪಾರ್ಶ್ವ ಬಾಹು
/
ವಿಕರ್ಣ
=
BC
/
AC
=
12
/
13
tan θ =
ಅಬಿಮುಖ ಬಾಹು
/
ಪಾರ್ಶ್ವ ಬಾಹು
=
AB
/
BC
=
5
/
12
ಮೂರು ಅಂಕದ ಪ್ರಶ್ನೆಗಳು
ಉತ್ತರ:
(sin θ + cosse θ)² + (cos θ + sec θ)²
= sin² θ + 2 x sin θ x cosec θ + cosec² θ + cos² θ + 2 x cos θ x sec θ + sec² θ
= sin² θ + 2 x sinθ x
{ cosec θ =
= sin² θ + 2 + cosec² θ + cos² θ + 2 + sec² θ
= sin² θ + cos² θ + 2 + 2 + cosec² θ + sec² θ
{ (1+tan²θ) = sec²θ, (1+cot²θ) = cosec²θ, sin² θ + cos² θ = 1 }
= 1 + 2 + 2 + 1 + cot² θ + 1 + tan² θ
= 1 + 2 + 2 + 1 + 1 + cot² θ + tan² θ
= 7 + cot² θ + tan² θ
(sin θ + cosse θ)² + (cos θ + sec θ)²
= sin² θ + 2 x sin θ x cosec θ + cosec² θ + cos² θ + 2 x cos θ x sec θ + sec² θ
= sin² θ + 2 x sinθ x
1
/
sin θ
+ cosec² θ + cos² θ + 2 x cosθ x
1
/
cos θ
+ sec² θ { cosec θ =
1
/
sin θ
sec θ =
1
/
cos θ
}= sin² θ + 2 + cosec² θ + cos² θ + 2 + sec² θ
= sin² θ + cos² θ + 2 + 2 + cosec² θ + sec² θ
{ (1+tan²θ) = sec²θ, (1+cot²θ) = cosec²θ, sin² θ + cos² θ = 1 }
= 1 + 2 + 2 + 1 + cot² θ + 1 + tan² θ
= 1 + 2 + 2 + 1 + 1 + cot² θ + tan² θ
= 7 + cot² θ + tan² θ
ಉತ್ತರ:
cos (A+B) = cos (60°+30°)
= cos 90°
= 0
cos A . cos B - sin A . sin B
= cos 60° . cos 30° - sin 60° . sin 30°
=
= 0
cos (A+B) = cos A . cos B - sin A . sin B
cos (A+B) = cos (60°+30°)
= cos 90°
= 0
cos A . cos B - sin A . sin B
= cos 60° . cos 30° - sin 60° . sin 30°
=
1
/
2
*
√3
/
2
-
√3
/
2
*
1
/
2
√3
/
4
-
√3
/
4
= 0
cos (A+B) = cos A . cos B - sin A . sin B
ಉತ್ತರ:
=
=
=
=
=
=
=
= 2 cosec θ
sin θ
/
1+cos θ
+
1+cosθ
/
sin θ
=
sin θ x sin θ + (1 + cos θ)(1 + cos θ)
/
sin θ (1 + cos θ)
=
sin² θ + (1 + cos θ)²
/
sin θ (1 + cos θ)
=
sin² θ + 1 + cos² θ + 2 x 1 x cos θ)
/
sin θ (1 + cos θ)
=
sin² θ + cos² θ + 1 + 2 cos θ)
/
sin θ (1 + cos θ)
=
1 + 1 + 2 cos θ)
/
sin θ (1 + cos θ)
=
2(1 + cos θ)
/
sin θ (1 + cos θ)
=
2
/
sin θ
= 2 cosec θ
ಉತ್ತರ:
=
=
=
= tan² θ
(1-tan² θ)
/
(cot² θ - 1)
=
1 -
sin² θ
/
cos² θ
/
cos² θ
/
sin² θ
- 1
=
cos² θ - sin² θ
/
cos² θ
/
cos² θ - sin² θ
/
sin² θ
=
sin² θ
/
cos² θ
= tan² θ
ಉತ್ತರ:
tan² θ - sin² θ
=
=
=
=
= tan² θ x sin² θ
tan² θ - sin² θ
=
sin² θ
/
cos² θ
- sin² θ =
(sin² θ - sin² θ cos² θ)
/
cos² θ
=
(sin² θ x (1-cos² θ))
/
cos² θ
=
sin²
/
cos² θ
x sin² θ = tan² θ x sin² θ
ಉತ್ತರ:
sin (90-θ) = cos θ
cosec (90-θ) = sec θ
cot (90-θ) = tan θ
=
=
=
=
=
=
= 1 + sin θ
sin (90-θ) = cos θ
cosec (90-θ) = sec θ
cot (90-θ) = tan θ
sin (90-θ)
/
(cosec (90-θ) - cot (90-θ) )
=
cos θ
/
sec θ - tan θ
=
cos θ
/
1
/
cos θ
-
sin θ
/
cos θ
=
cos θ
/
1 - sin θ
/
cos θ
=
cos² θ
/
1 - sin θ
=
1 - sin² θ
/
1 - sin θ
=
(1 - sin θ)(1 + sin θ)
/
1 - sin θ
= 1 + sin θ