Chapter 8 Introduction to Trigonometry and its Equation NCERT Exemplar Solutions Exercise 8.2 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 8 Introduction to Trigonometry and its Equation Exercise 8.2 
Book Name  NCERT Exemplar for Class 10 Maths 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 8.2 Solutions
Short Answer Questions with Reasoning:
Write ‘True’ or ‘False’ and justify your answer in each of the following:
1. tan 47°/cot 43° = 1
Solution
True
As, tan (90°  Î¸) = cot Î¸
tan 47°/cot 43° = [tan(90  43°)]/cot 43°
= cot 43°/cot 43°
= 1
2. The value of the expression (cos^{2} 23°  sin^{2} 67°) is positive.
Solution
False
As,
(a^{2}  b^{2} ) = (a + b)(a  b)
cos^{2} 23°  sin^{2} 67° =(cos 23°+sin 67°)(cos 23°sin 67°)
= [cos 23°+sin(90°23°)] [cos 23°sin(90°23°)]
= (cos 23°+cos 23°)(cos 23°cos 23°)
(as, sin(90°Î¸) = cos Î¸)
= (cos 23°+cos 23°).0
= 0,
And 0 is neither positive nor negative.
3. The value of the expression (sin 80°  cos 80°) is negative .
Solution
False
As,
sin Î¸ increases when 0° ≤ Î¸ ≤ 90°
Therefore,
(sin 80°cos 80°) = (increasing valuedecreasing value)
= a positive value.
So,
(sin 80° cos 80°) > 0.
4. √[(1  cos^{2} Î¸)sec^{2} Î¸ = tanÎ¸
Solution
True
5. If cos A + cos^{2} A = 1, then sin^{2} A + sin^{4} A = 1
Solution
True
As given in the question,
cos A + cos^{2} A = 1
⇒ cos A = 1  cos^{2} A
Also,
sin^{2} Î¸+cos^{2} Î¸ = 1
⇒ sin^{2} Î¸ = 1 cos^{2} Î¸
So,
cos A = sin^{2} A ...(1)
Squaring L.H.S and R.H.S,
cos^{2} A = sin^{2} A ...(2)
Adding equations (1) and (2),
We get
sin^{2} A + sin^{4} A= cos A + cos^{2} A
So,
sin^{2} A+ sin^{4} A = 1
6. (tanÎ¸ + 2)(2tanÎ¸ + 1) = 5tanÎ¸ + sec^{2} Î¸
Solution
False
L.H.S
= (tan Î¸ + 2) (2 tanÎ¸+1)
= 2 tan^{2} Î¸ + tanÎ¸ + 4 tanÎ¸ + 2
= 2 tan^{2} Î¸+5 tanÎ¸ + 2
As,
sec^{2} Î¸ – tan^{2} Î¸ = 1,
we get,
tan^{2} Î¸ = sec^{2} Î¸  1
= 2(sec^{2} Î¸1) + 5 tanÎ¸+2
= 2 sec^{2} Î¸  2 + 5 tan Î¸+2
= 5 tanÎ¸+ 2 sec^{2} Î¸
So,
L.H.S ≠ R.H.S
7. If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is also increasing.
Solution
False
To understand the fact of this question, consider the following example
(i) A tower 2√3 m high casts a shadow 2m long on the ground, then the Sun's elevation is 60°
In Î”ABC,
tan Î¸ = BC/AB
⇒ tan Î¸ = 2√3/2
⇒ tan Î¸ = √3
⇒ tan Î¸ = 60°
(ii) The A same height of tower casts a shadow 4m more from preceding point, then the Sun's elevation is 30°
In Î”PBC,
tanÎ¸ = CB/PB
⇒ tanÎ¸ = CB/(PA + AB)
⇒ tanÎ¸ = 2√3/(4+2)
⇒ tanÎ¸ = 2√3/6
⇒ tanÎ¸ = 1/√3
⇒ tanÎ¸ = 30°
Therefore, we conclude from above two examples that if the length of the shadow of a tower is increasing, then the angle of elevation of the Sun is decreasing.
8. If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.
Solution
False
We can observe that, a man standing on a platform at point P, 3 m above the surface of a lake observes a cloud at point C.
Let the height of the cloud from the surface of the platform is h.
In MPC,
Then,
a + 1/a = 2 + 1/2
= 5/2
If 2sinÎ¸ = a + 1/a , then,
2sinÎ¸ = 5/2
⇒ sinÎ¸ = 5/4
= 1.25
Which is not possible.
⇒ a^{2} + b^{2} > 2ab
⇒ (a^{2} +b^{2} )/2ab > 1
cosÎ¸ = cosÎ¸ > 1 [as, cos Î¸ = (a^{2} +b^{2} )/2ab]
Which is not possible. [as, 1 ≤ cosÎ¸ ≤ 1]
So, cosÎ¸ ≠ (a^{2} +b^{2} )/2ab
Case (i): Let us take the height of the tower is h In Î”ABC, let BC = x m
In ABC,
tan 30° = AC/BC
1/√3 = h/x ...(i)
Case (ii) : By condition, the height of the tower is doubled, PR = 2h,
In Î”PQR,
tanÎ¸ = PQ/QR
⇒ tanÎ¸ = 2h/x
⇒ tanÎ¸ = 2/x × x/√3 ...(from (i))
⇒ tanÎ¸ = 2/1 × 1/√3
⇒ tanÎ¸ = 1.15
⇒ Î¸ = tan^{1} (1.15)
which is less than 60°
So, the required angle is not doubled.
Let us take the height of a tower be h and the distance of the point of observation from its foot is x.
In ABC,
tan Î¸_{1} = AC/BC
= h/x ...(i)
Case (ii) :
Now, the height of a tower increased by 10%
= h + 10% of h
= h + h× 10/100
= 11h/100
and the distance of the point of observation from its foot
= x + 10% of x
= 11x/10
In PQR,
tan Î¸_{2} = PR/QR
= (11h/10)/(11x/10)
= 4/x ...(ii)
From (i) and (ii),
tanÎ¸_{1} = tanÎ¸_{2}