Evaluate sin square 63 + sin square 27 | Evaluate sin 25 degree cos 65 degree [Q3 both parts]

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Exercise 8.4 question 3 class 10
Both parts solved [Important Question]

Evaluate:
1. (〖sin〗^(2 ) 63° +〖sin 〗^(2 ) 27° )/(〖cos〗^2 17°+ 〖cos〗^(2 ) 73°)
or
sin square 63 degree + sin square 27 degree by cos square 17 + cos square 73 [0:12]
2. sin 25°cos65 °+cos⁡ 25°sin⁡ 65°
or
sin 25 degree cos 65 degree + cos 25 degree sin 65 degree [3:55]

Hint 💡
Trigonometric identity:
cos^2 A + sin^2 A = 1
Complementary angles:
sin (90° - A ) = cos A
cos (90° - A) = sin A

Note📝
The above question is taken from NCERT book ch-8 exercise 8.4 question 3 Trigonometry.

Read it 📖
Identity:
An equation is called an identity when it is true for all values of the variables involved.

Trigonometric identity :
Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved.

Practice👩‍🏫:
Write all the other trigonometric ratios of angle A in terms of sec A.

Trigonometric identities

Trigonometry playlist [latest]

Revise✍🏻:
Quadratic Equations

Polynomials

Real Numbers

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#ncert #class10maths #trigonometry
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RHAcademy_
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Jo question puch rahi hu uska answer hi mil Raha hai 😣😣😣😣😣😩😩😩😩😩😫😫😫😫😫😫

Smishra
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Kindly note.
[0:40] sin ( 90 - A) = cos A
Rather
sin A = cos ( 90 - A)
where A is 63 degree

[2:06] cos (90-A) = sin A
Rather
sin A = cos ( 90-A)
where A is 17 degree

[4:46] cos (90-A) = sin A
Rather
sin A = cos ( 90-A)
where A is 25 degree

[5:47] sin ( 90 - A) = cos A
Rather
sin A = cos ( 90 - A)
where A is 65 degree

*Do not make the same mistake* 😌
Apology for the errors 🙏

RHAcademy_
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Mam please use hindi to learn easy. It is not possible to understand English to everyone,

AYUSHSHARMA-qghj
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English aja ra he beach me accha se nahi dik raha he

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