SOLUTION:

Soln. : In right angle ΔABC, we have AB = 24 cm, BC = 7 cm

∴ Using Pythagoras theorem, AC² = AB² + BC²
⇒ AC² = 24² + 7² = 576 +49 = 625 = 25²
⇒ AC = 25 cm
(i)
(ii) ,

SOLUTION:

Soln. : In right angle ΔPQR
Using the Pythagoras theorem, we get
QR² = PR² - PQ²
⇒ QR² = 13² - 12² = (13 - 12)(13 + 12)
= 1 × 25 = 25
cm
Now, ,

SOLUTION:

Soln. : Let us consider, the right ΔABC, we have
Perpendicular = BC and

Hypotenuse = AC

Let BC = 3k and AC = 4k

= (4k)² - (3k)² = (4k - 3k)(4k + 3k) = k(7k) = 7k²

SOLUTION:

Soln. : In the right triangle ABC, we have
15 cot A = 8




∴ AB = 8k and BC = 15k
Now, using Pythagoras theorem, we get
AC² = AB² + BC²
= (8k)² + (15k)² = 64k² + 225k² = 289k² = (17k)²
⇒ AC =

SOLUTION:

Soln. : In right ΔABC, ∠ B = 90°

Let ∠ A = θ and sec θ =

⇒ AC = 13k and AB = 12k
∴ Using Pythagoras theorem, we get
BC² = AC² - AB²
⇒ BC² = (13k)² - (12k)²
= (13k - 12k) (13k + 12k)
= k(25k) = 25k² = (5k)²

SOLUTION:

Soln. : Let us consider a right ΔABC, ∠ C = 90°
Now, cos A =

And
Since, cos A = cos B

Now, in the ΔABC, two sides AC and BC are equal. [Proved above]
∴ Their opposite angles are also equal.
∴ ∠ A = ∠ B.

SOLUTION:

Soln. : In right ΔABC, ∠ B = 90° and ∠ A = θ
.......(1) [Given]
But in right ΔABC, cot θ = .......(2)

From (1) and (2),
⇒ AB = 7k and BC = 8k
∴ AC² = AB² + BC² = (7k)² + (8k)²
(Pythagoras theorem)



Now,
(i)

(ii)

SOLUTION:

Soln. : In right angled ΔABC, ∠ B = 90°
∴ For ∠ A, we have:
Base = AB and perpendicular = BC. Also
Hypotenuse = AC

∴ 3 cot A = 4
∴ cot A = 4/3 ...(1)
But cot A ...(2)
∴ From (1) and (2),
⇒ AB = 4k and BC = 3k
∴ Using Pythagoras theorem, we get
AC² = AB² + BC²
⇒ AC² = (4k)² + (3k)²

Now,


Now, to check the given equation,
L.H.S. =
........(i)
R.H.S. = cos²A - sin²A
.........(ii)
From (i) and (ii), we have L.H.S. = R.H.S.

SOLUTION:

Soln. : In right ΔABC, ∠ B = 90°
For ∠ A, we have
Base = BC, Perpendicular = AB,
Hypotenuse = AC
.......(1) (Given)
∴ tan A ........(2)
From (1) and (2), we have

⇒ AB = and BC = 1k
Now, using Pythagoras theorem, we get
AC² = AB² + BC²


Now, ,

Again, for ∠ C, we have
Base = BC, Perpendicular = AB
and Hypotenuse = AC
,

(i)

(ii)

SOLUTION:

Soln. : In right ΔPQR, Q = 90°

PR + QR = 25 cm and PQ = 5 cm
Let QR = x cm ⇒ PR = (25 - x)
∴ By Pythagoras theorem, we have
PR² = QR² + PQ²
⇒ (25 - x)² = x² + 5²
⇒ 625 - 50x + x² = x² +25
⇒ - 50x = - 600
⇒ i.e., RQ = 12 cm
⇒ RP = 25 - 12 = 13 cm
Now, , ,

SOLUTION:

Soln. : (i) False [∵ A tangent of an angle is ratio of sides other than hypotenuse, which may be equal or unequal to each other.]
(ii) True [∵ cos A is always less than 1]
i.e., sec A will always be greater
than 1
(iii) False [∵ 'cosine A' is abbreviated as 'cos A']
(iv) False ['cot A' is a single and meaningful term whereas 'cot' alone has no meaning.]
(v) False [ is greater than 1 and sinθ cannot be greater than 1]

SOLUTION:

Soln. : (i) sin 60° cos 30° + sin 30° cos 60°




(ii) 2tan² 45° + cos² 30° - sin² 60°




(iii)






(iv)

and cot 45° = 1





(v)

SOLUTION:

Soln. : (i) (A)


(ii) (D) :

(iii) (A) : When A = 0, then
sin 2A = sin 2(0°) = sin 0° = 0,
2 sin A = 2 sin 0° = 2 × 0 = 0
i.e., sin 2A = 2sin A for A = 0°
(iv) (C) :

SOLUTION:

Soln. : We have,
tan 60° , tan 30° = ......(1)
Also tan (A + B) and tan (A - B) (Given)
.......(2)
From (1) and (2), we get A + B = 60° and
A - B = 30°
On adding A + B and A - B, we get 2A = 90°
⇒ A = 45°
On subtracting A + B and A - B, we get 2B = 30°
⇒ B = 15°

SOLUTION:

Soln. : (i) False : Let us take A = 30° and B = 60°, then L.H.S = sin (30° + 60°) = sin 90° = 1
R.H.S.= sin 30° + sin 60°

∴ L.H.S. ≠ R.H.S.
(ii) True : Since the values of sin θ increases from 0 to 1 as θ increases from 0 to 90°.
(iii) False : Since the value of cos θ decreases from 1 to 0 as θ increases from 0 to 90°.
(iv) False : Let us take θ = 30°

⇒ sin 30° ≠ cos 30°
(v) True : We have cot 0° = not defined

SOLUTION:

Soln. : (i)
⇒ sin 18° = sin (90° - 72°) = cos 72°
[∵ sin (90° - A) = cos A]
∴
(ii)
tan 26° = tan (90° - 64°) = cot 64° [∵ tan (90° - A) = cot A]

(iii) cos 48° - sin 42°
cos 48° = cos (90° - 42°) = sin 42° [∵ cos (90° - A) = sin A]
∴ cos 48° - sin 42° = sin 42° - sin 42° = 0
(iv) cosec 31° - sec 59°
cosec 31° = cosec (90° - 59°) = sec 59° [∵ cosec (90° - A) = sec A]
∴ cosec 31° - sec 59° = sec 59° - sec 59° = 0

SOLUTION:

Soln. : (i) tan 48° tan 23° tan 42° tan 67° = 1
L.H.S. = tan 48° tan 23° tan 42° tan 67°
= tan (90° - 42°) tan 23° tan 42° tan (90° - 23°)
[Q tan (90° - A) = cot A]
= cot 42° tan 23° tan 42° cot 23°



∴ L.H.S. = R.H.S.
(ii) cos 38° cos 52° - sin 38° sin 52° = 0
L.H.S. = cos 38° cos 52° - sin 38° sin 52°
= cos 38° cos (90° - 38°) - sin 38° sin (90° - 38°)
= cos 38° sin 38° - sin 38° cos 38°
[∵ sin (90° - A) = cos A and cos (90° - A)= sin A]
= 0 = R.H.S.
∴ L.H.S. = R.H.S.

SOLUTION:

Soln. : Since tan 2A = cot (A - 18°)
Also tan (2A) = cot (90° - 2A)
[∵ tan θ = cot (90° - θ)]
⇒ A - 18° = 90° - 2A
⇒ A + 2A = 90° + 18°
⇒ 3A = 108° ⇒

SOLUTION:

Soln. : tan A = cot B and cot B = tan (90° - B)
∴ A = 90° - B ⇒ A + B = 90°
[∵ tan (90° - θ) = cot θ]

SOLUTION:

Soln. : ∵ sec 4A = cosec (A - 20°)
sec (4A) = cosec (90° - 4A)
[∵ cosec (90° - θ) = sec θ]
∴ A - 20° = 90° - 4A
⇒ A + 4A = 90° + 20°
⇒ 5A = 110° ⇒ A =

SOLUTION:

Soln. : Since, sum of the angles of ΔABC is
A + B + C = 180°
∴ B + C = 180° - A
Dividing both sides by 2, we get


[∵ sin (90° - θ) = cos θ]

SOLUTION:

Soln. : Since sin 67° = sin (90° - 23°) = cos 23°
[∵ sin (90° - θ ) = cos θ]
Also, cos 75° = cos (90° - 15°) = sin 15°
[∵ cos (90° - θ) = sin θ]
∴ sin 67° + cos 75° = cos 23° + sin 15°

SOLUTION:

Soln. : (a) sin A

(b)

(c)

SOLUTION:

Soln. : (i) sin A =

(ii)
(iii)
(iv)

(v)

SOLUTION:

Soln. : (i)
∵ sin 63° = sin (90° - 27°) = cos 27°
⇒ sin² 63° = cos² 27°
cos² 73° = cos² (90° - 17°) = sin² 17°
= 1
[∵ cos²A + sin²A = 1]
(ii) sin 25° cos 65° + cos 25° sin 65°
∵ sin 25° = sin (90° - 65°) = cos 65°
[∵ sin (90° - A) = cos A]
cos 25° = cos (90° - 65°) = sin 65°
[∵ cos (90° - A) = sin A]
∴ sin 25° cos 65° + cos 25° sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= (cos 65°)² + (sin 65°)²
= cos² 65° + sin² 65° = 1 [∵ cos² A + sin² A = 1]

SOLUTION:

Soln. : (i) (B) : Since, 9 sec² A - 9 tan² A
= 9 (sec² A - tan² A) = 9 (1) = 9
[∵ tan² A + 1 = sec² A ⇒ sec² A - tan²A = 1]
(ii) (C) : Here, (1 + tan θ + sec θ) (1 + cot θ - cosec θ)
= (1 + tan θ + sec θ)
= (1 + tan θ + sec θ)





= 2
(iii) (D) : We have : (sec A + tan A) (1 - sin A)


= cos A
(iv) (D) : Here,
= tan² A

SOLUTION:

Soln. : (i) L.H.S = (cosec θ - cot θ)²

[∵ sin² θ = 1 - cos² θ]

[∵ 1 - cos² θ = (1 - cos θ) (1 + cos θ)]

Hence, L.H.S. = R.H.S.
(ii) L.H.S.



[∵ cos² A + sin² A = 1]


= R.H.S
Hence, L.H.S. = R.H.S.
(iii) L.H.S.







[∵ sin² θ + cos² θ = 1]

= cosec θ sec θ + 1 = 1 + sec θ·cosec θ
= R.H.S.
(iv)

[Multiplying and dividing by (1 - cos A)]

= R.H.S. [∵ 1 - cos² A = sin² A]
(v) L.H.S.
[Dividing each term of num. and den. by sin A]

[Multiplying and dividing by (cot A + cosec A)]



[∵ cot² A - cosec² A = - 1]
= cot A + cosec A = R.H.S.
(vi) L.H.S.
[Multiplying and dividing by (cot A + cosec A)]
[∵ (1 - sin A) (1 + sin A)
= 1 - sin² A]
[∵ 1 - sin² A = cos² A]

= sec A + tan A = R.H.S.
(vii)



(viii) L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= sin² A + cosec² A + 2 sin A·cosec A + cos² A
+ sec² A + 2 cos A·sec A
= (sin² A + cos² A) + cosec² A + sec² A + 2 + 2
[sin A·cosec A = 1 and sec A·cos A = 1]
= 1 + cosec² A + sec² A + 4
[∵ sin² A + cos² A = 1]
= 5 + (1 + cot² A) + (1 + tan² A)
[∵ cosec² A = 1 + cot² A and sec² A = 1 + tan² A]
= 7 + cot² A + tan² A = R.H.S
(ix) L.H.S. = (cosec A - sin A) (sec A - cos A)


[∵ 1 - sin² A = cos² A
and 1 - cos² A = sin² A]
= sin A·cos A

[∵ 1 = sin² A + cos² A]
[Dividing num. and den. by sin A cos A]

(x)

= tan² A = R.H.S. ...(1)



= (-tan A)² = tan²A = R.H.S. ...(2)
∴ From (1) and (2), we get