Exercise 8.3 class 10 maths | Trigonometry Questions Explained | NCERT Solutions Class 10

Описание: Exercise 8.3 class 10 maths | Trigonometry Questions Explained | NCERT Solutions Class 10📚 Class 10 Maths Exercise 8.3 (Trigonometry) Full Solution
In this video, we will solve all the questions of NCERT Class 10 Maths Exercise 8.3 step by step.
This video will help you understand Trigonometry concepts, formulas, and problem-solving techniques in an easy way.

✅ Topics Covered:

Trigonometry Class 10 Important Questions

Exercise 8.3 NCERT Complete Solutions

Step by Step Explanation for Board Exam Preparation


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answers of Maths NCERT Class 10 Chapter 8 – Introduction to Trigonometry Exercise 8.4
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution:

To convert the given trigonometric ratios in terms of the cot functions, use trigonometric formulas.

We know that,

cosec2A – cot2A = 1

cosec2A = 1 + cot2A

Since the cosec function is the inverse of the sin function, it is written as

1/sin2A = 1 + cot2A

Now, rearrange the terms; it becomes

sin2A = 1/(1+cot2A)

Now, take square roots on both sides; we get

sin A = ±1/(√(1+cot2A)

The above equation defines the sin function in terms of the cot function

Now, to express the sec function in terms of the cot function, use the formula

sin2A = 1/ (1+cot2A)

Now, represent the sin function as the cos function

1 – cos2A = 1/ (1+cot2A)

Rearrange the terms,

cos2A = 1 – 1/(1+cot2A)

⇒cos2A = (1-1+cot2A)/(1+cot2A)

Since the sec function is the inverse of the cos function,

⇒ 1/sec2A = cot2A/(1+cot2A)

Take the reciprocal and square roots on both sides, and we get

⇒ sec A = ±√ (1+cot2A)/cotA

Now, to express the tan function in terms of the cot function

tan A = sin A/cos A and cot A = cos A/sin A

Since the cot function is the inverse of the tan function, it is rewritten as

tan A = 1/cot A

2. Write all the other trigonometric ratios of ∠A in terms of sec A.

Solution:

Cos A function in terms of sec A:

sec A = 1/cos A

⇒ cos A = 1/sec A

sec A function in terms of sec A:

cos2A + sin2A = 1

Rearrange the terms.

sin2A = 1 – cos2A

sin2A = 1 – (1/sec2A)

sin2A = (sec2A-1)/sec2A

sin A = ± √(sec2A-1)/sec A

cosec A function in terms of sec A:

sin A = 1/cosec A

⇒cosec A = 1/sin A

cosec A = ± sec A/√(sec2A-1)

Now, tan A function in terms of sec A:

sec2A – tan2A = 1

Rearrange the terms.

⇒ tan2A = sec2A – 1

tan A = √(sec2A – 1)

cot A function in terms of sec A:

tan A = 1/cot A

⇒ cot A = 1/tan A

cot A = ±1/√(sec2A – 1)

Hence proved.

(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A

L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2

It is of the form (a+b)2, expand it

(a+b)2 =a2 + b2 +2ab

= (sin2A + cosec2A + 2 sin A cosec A) + (cos2A + sec2A + 2 cos A sec A)

= (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A

= 1 + 2 + 2 + 2 + tan2A + cot2A

= 7+tan2A+cot2A = R.H.S.

Therefore, (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A

Hence proved.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

First, find the simplified form of L.H.S.

L.H.S. = (cosec A – sin A)(sec A – cos A)

Now, substitute the inverse and equivalent trigonometric ratio forms.

= (1/sin A – sin A)(1/cos A – cos A)

= [(1-sin2A)/sin A][(1-cos2A)/cos A]

= (cos2A/sin A)×(sin2A/cos A)

= cos A sin A

Now, simplify the R.H.S.

R.H.S. = 1/(tan A+cotA)

= 1/(sin A/cos A +cos A/sin A)

= 1/[(sin2A+cos2A)/sin A cos A]

= cos A sin A

L.H.S. = R.H.S.

(cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

Hence proved.

(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A

L.H.S. = (1+tan2A/1+cot2A)

Since the cot function is the inverse of the tan function,

= (1+tan2A/1+1/tan2A)

= 1+tan2A/[(1+tan2A)/tan2A]

Now cancel the 1+tan2A terms, and we get

= tan2A

(1+tan2A/1+cot2A) = tan2A

Similarly,

(1-tan A/1-cot A)2 = tan2A

Hence proved.