Introduction to Trigonometry
Mathematics Class Tenth
Solution of NCERT Exercise 8.1 part3
Question (8) If 3cot A = 4, check whether 1 – tan2A/1 + tan2A = cos2A – sin2A or not
Solution:
Given, 3 cot A = 4
⇒ cot A = 4/3 --------- (i)
We know that, cot A = Base(b)/Perpendicular(p)
Thus, after substituting the value of cot A from equation (i) we get
⇒ 4/3 = Base(b)/Perpendicular(p)
Thus, Base (b) = 4 k, and Perpendicular(p) = 3 k
According to Pythagoras Theorem, we know that,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Thus, by substituting values of b (base) and p (perpendicular) we get
(Hypotenuse)2 = (3 k)2 + (4 k)2
⇒ (Hypotenuse)2 = 9 k 2 + 16 k2
⇒ (Hypotenuse)2 =25 k2
⇒ Hypotenuse (h) = 25 k2
⇒ Hypotenuse (h) = 5 k
Now, we have, Base (b) = 4 k, Perpendicular (p) = 3 k and Hypotenuse(h) = 5 k
Calculation of the trigonometric ratio of tan A
Now, we know that, tan A = Perpendicular(p)/Base(b)
By substituting values of p (perpendicular) and b (base), we get
tan A = 3 k/4 k
⇒ tan A = 3/4 - - - - - (ii)
Calculation of the trigonometric ratio of cos A
Now, we know that, cos A = Base(b)/Hypotenuse(h)
Thus, by substituting values of b(base) and h (hypotenuse), we get
cos A = 4 k/5 k
⇒ cos A = 4/5 - - - - - (iii)
Calculation of the trigonometric ratio of sin A
Now, we know that, sin A = Perpendicular(p)/Hypotenuse(h)
Thus, by substituting values of p(perpendicular) and h (hypotenuse), we get
sin A = 3 k/5 k
⇒ sin A = 3/5 - - - - - (iv)
Now, LHS = 1 – tan2A/1 + tan2A
After substituting value of tan A from equation (ii), we get
1 – (3/4)2/1 + (3/4)2
= 1 – 9/16/ 1 + 9/16
= 16 – 9/16/16 + 9/16
= 7/16/25/16
= 7/16 × 16/25
⇒ LHS = 7/25
Now, RHS = cos2A – sin2A
After substituting values of cos A and sin A from equation (iii) and (iv), we get
=(4/5)2 – (3/5)2
= 16/25 – 9/25
= 16 – 9/25
⇒ RHS = 7/25
Thus, LHS = RHS Proved
i.e. 1 – tan2A/1 + tan2A = cos2A – sin2A Proved
Alternate Method to solve the question
If 3cot A = 4, check whether 1 – tan2A/1 + tan2A = cos2A – sin2A or not
Given, 3 cot A = 4
∴ cot A = 4/3 - - - - - (i)
Calculation of the trigonometric ratio of tan A using given value of cot A
Now, we know that, tan A = 1/cot A
Thus, by substituting the value of cot A from equation (i), we get
tan A = 1/4/3
⇒ tan A = 3/4 - - - - (ii)
Calculation of the trigonometric ratio of cosec A using given value of cot A
Now, we know that, cot2 A = cosec2A – 1
After substituting value of cot A from equation (i), we get
(4/3)2 = cosec2A – 1
⇒ cosec2A =(4/3)2 + 1
⇒ cosec2A = 16/9 + 1
⇒ cosec2A = 16 + 9/9
⇒ cosec2 A = 25/9
⇒ cosec A = 25/9
⇒ cosec A = 5/3
Calculation of the trigonometric ratio of sin A using given value of cosec A
Now, we know that, sin A = 1/cosec A
⇒ sin A = 1/5/3
⇒ sin A = 3/5 - - - - - (iii)
Calculation of the trigonometric ratio of cos A using given value of sin A
We know that, cos2 A = 1 – sin2A
After substituting value of sin A from equation (iii), we get
cos2 A = 1 – (3/5)2
⇒ cos2 A = 1 – 9/25
⇒ cos2A = 25 – 9/25
⇒ cos A = 16/25
⇒ cos A = 4/5 - - - - - (iv)
Now, LHS = 1 – tan2A/1 + tan2A
After substituting value of tan A from equation (ii), we get
1 – (3/4)2/1 + (3/4)2
= 1 – 9/16/ 1 + 9/16
= 16 – 9/16/16 + 9/16
= 7/16/25/16
= 7/16 × 16/25
⇒ LHS = 7/25
Now, RHS = cos2A – sin2A
After substituting values of cos A and sin A from equation (iii) and (iv), we get
=(4/5)2 – (3/5)2
= 16/25 – 9/25
= 16 – 9/25
⇒ RHS = 7/25
Thus, LHS = RHS Proved
i.e. 1 – tan2A/1 + tan2A = cos2A – sin2A Proved
Question (9) In triangle ABC, right angled at B, if tan A = 1/√3, find the value of
(i) sin A . cos C + cos A . sin C
(ii) cos A . cos C – sin A . sin C
Solution:
Let Δ ABC is a right-angle triangle, in which ∠B = 90o
Here, since the acute ∠A is to be considered, thus for acute ∠ A,
the side opposite to the ∠ B = AC = Hypotenuse (h)
The side opposite to the ∠ A = BC = Perpendicular (p)
And, the side adjacent to the ∠ A = AB = Base (b)
Given, tan A = 1/√3
We know that, tan A = Perpendicular (p)/Base (b)
Thus, tan A = Perpendicular(p)/Base(b) = 1/3
∴ Perpendicular (p) = BC = k
And Base (b) = AB = √3 k
Calculation of the Hypotenuse
According to Pythagoras Theorem, we know that,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
After substituting values of Perpendicular(p) and Base(b), we get
⇒ [Hypotenuse(h)]2 = k2 + (√3 k)2
⇒ [Hypotenuse(h)]2 = k2 + 3 k2
⇒ Hypotenuse(h) = 4 k2
⇒ Hypotenuse(h) = AC = 2 k
Calculation of the sinA using the value of tanA
Now, we know that sin A = BC/AC = Perpendicular (p)/Hypotenuse (h)
After substituting values of perpendicular (p) and hypotenuse (h), we get
sin A = k/2 k
⇒ sin A = 1/2 - - - - - (i)
Calculation of the cos A using the value of tan A
We know that, cos A = AB/AC
After substituting values of base(b) and hypotenuse(h), we get
cos A = √3 k/2 k
⇒ cos A = √3/2 - - - - - (ii)
Now in the given Δ ABC, for the acute ∠ C
the side opposite to the right ∠ B = AC = Hypotenuse (h) = 2k
The side opposite to the ∠ C = AB = Perpendicular (p) = 3k
And, the side adjacent to the ∠ C = BC = Base (b) = k
Calculation of the sin C using the values given for the right angle triangle ABC
Now, we know that, sin C = Perpendicular (p)/Hypotenuse (h)
⇒ sin C = AB/AC
After substituting values of AB and AC, we get
sin C = √3 k/2 k
⇒ sin C = √3/2 - - - - - (iii)
Calculation of the cos C using the values given for the right angle triangle ABC
Now, we know that, cos C = Base (b)/Hypotenuse (h)
BC/AC
After substituting values of Base (BC) and Hypotenuse (AC), we get
cos C = k/2 k
⇒ cos C = 1/2 - - - - (iv)
Now,
(i) sin A . cos C + cos A . sin C
After substituting values of sin A, cos A, sin C and cos C from equation (i), (ii), (iii) and (iv), we get
1/2 × 1/2 + √3/2 × √3/2
= 1/4 + 3/4
= 1 + 3/4
= 1/4
Thus, sin A . cos C + cos A . sin C = 1/4 Answer
(ii) cos A . cos C – sin A . sin C
After substituting values of sin A, cos A, sin C and cos C from equation (i), (ii), (iii) and (iv), we get
√3/2 × 1/2 – 1/2 × √3/2
= √3/4 – √3/4
= 0
Thus, cos A . cos C – sin A . sin C = 0 Answer
Question (10) In Δ PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
Let, ABC is the given triangle
∠ Q = 90o (As given in the question)
Thus, in this right angle triangle, for the acute ∠ P
The side opposite to right ∠Q = PR = Hypotenuse (h)
And, side adjacent to the acute ∠P = Base (b) = PQ
And, the side opposite to the acute ∠ P = Perpendicular (p)
As given in question,
∠ Q = 90o
And, PQ = Base (b) 5 cm
And, PR + QR = 25 cm
∴ QR = Perpendicular (p) = 25 cm – PR - - - - (i)
Now, according to Pythagoras Theorem,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
⇒ PR2 = QR2 + PQ2
⇒ PR2 = (25 cm – PR)2 + (5 cm)2
[∵ from equation (i) QR = 25 – PR]
⇒ PR2 = 625 cm2 + PR2 – 2 × 25 cm × PR cm + 25 cm2
⇒ PR2 = 625 cm2 + PR2 – PR × 50 cm + 25 cm2
⇒ PR2 = 650 cm2 + PR2 – PR × 50 cm
⇒ PR2 – PR2 = 650 cm2 – PR × 50 cm
⇒ 0 = 650 cm2 – PR × 50 cm
⇒ PR × 50 cm = 650 cm2
⇒ PR = 650 cm2/50 cm
⇒ PR = 13 cm
Now, by substituting value of PR in equation (i), we get
QR = 25 cm – 13 cm
⇒ QR = 12 cm
Thus, now we have
PQ = Base(b) = 5 cm,
QR = Perpendicular (p) = 12 cm,
And, PR = Hypotenuse (h) = 13 cm
Calculation of sinP using the values given for the right angle ABC
Now, We know that, sin P = Perpendicular (p)/Hypotenuse (h)
⇒ sin P = QR/PR
Substituting values of QR and PR we get
sin P = 12 cm/13 cm
⇒ sin P = 12/13
Calculation of cosP using the values given for the right angle ABC
We know that, cos P = Base (b)/Hypotenuse (h)
⇒ cos P = PQ/PR
After substituting values of PQ and PR we get
cos P = 5 cm/13 cm
⇒ cos P = 5/13
Calculation of tanP using the values given for the right angle ABC
Again, we know that, tan P = Perpendicular (p)/Base (b)
⇒ tan P = QR/PQ
After substituting values of QR and PQ we get
tan P = 12 cm/5 cm
⇒ tan P = 12/5
Thus, sin P = 12/13, cos P = 5/13 and tan P = 12/5 Answer
Question (11) State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
Answer: False
Justification: The value of tan 45o is equal to 1 and hence the statement "The value of tan A is always less than 1" is false.
(ii) sec A = 12/5 for some value of angle A.
Answer: True
Justification:
We know that, sec A = h/b
Here, given, sec A = 12/5
Thus, sec A = 12/5 = h/b
Thus, h = 12 k, and b = 5 k
We know that, According to Pythagoras Theorem,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
⇒ (12 k)2 = p2 + (5 k)2
⇒ p2 = 144 k2 – 25 k2
⇒ p2=119 k2
⇒ p2 = 119 k2
⇒ p = 10.9
Since, here hypotenuse (12) is biggest among all the sides and sum of perpendicular and base is more than hypotenuse, thus, such triangle will exist.
Thus, the statement "sec A = 12/5 for some value of angle A" is true.
(iii) cos A is the abbreviation of used for the cosecant of angle A.
Answer: False
Justification: "cos A" is the abbreviation used for "cosine of angle A" and "cosec A" is the abbreviation used for "cosecant of angle A".
Thus the given statement "cos A is the abbreviation of used for the cosecant of angle A" is false
(iv) cot A is the product of cot and A.
Answer: False
Justification: "cot A" is not the product of cot and A, rather "cot A" is the "cotangent of angle A".
Thus, the given statement "cot A is the product of cot and A" is false
(v) sin θ = 4/3 for some angle θ.
Answer: False
Justification: The value of sin θ is always between or equal to 1 and 0. Here, the value of sin θ is equal to 4/3 which is more than 1. Thus, given statement "sin θ = 4/3 for some angle θ" is false.
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