In Î”*ABC* , right-angled at *B* , *AB* = 24 cm,

*BC* = 7 cm. Determine :

(i) sin *A* , cos *A* (ii) sin *C* , cos *C*

**Soln. :** In right angle Î”*ABC*, we have *AB* = 24 cm, *BC* = 7 cm

âˆ´ Using Pythagoras theorem, *AC*^{2} = *AB*^{2} + *BC*^{2}

â‡’ *AC*^{2} = 24^{2} + 7^{2} = 576 +49 = 625 = 25^{2}

â‡’ *AC* = 25 cm

(i)

(ii) ,

In the figure, find tan *P* - cot *R* .

**Soln. :** In right angle Î”*PQR*

Using the Pythagoras theorem, we get
*QR*^{2 }= *PR*^{2} - *PQ*^{2}

â‡’ *QR*^{2} = 13^{2} - 12^{2} = (13 - 12)(13 + 12)

= 1 Ã— 25 = 25

cm

Now, ,

If calculate cos *A* and tan *A* .

**Soln. :** Let us consider, the right Î”*ABC*, we have

Perpendicular = *BC* and

Hypotenuse = *AC*

Let *BC* = 3*k* and *AC* = 4*k*

= (4*k*)^{2} - (3*k*)^{2 }= (4*k* - 3*k*)(4*k* + 3*k*) = *k*(7*k*) = 7*k*^{2}

Given 15 cot *A* = 8, find sin *A* and sec *A* .

**Soln. :** In the right triangle *ABC*, we have

15 cot *A* = 8

âˆ´ *AB* = 8*k* and *BC* = 15*k*

Now, using Pythagoras theorem, we get
*AC*^{2} = *AB*^{2} + *BC*^{2}

= (8*k*)^{2} + (15*k*)^{2} = 64*k*^{2} + 225*k*^{2} = 289*k*^{2} = (17*k*)^{2}

â‡’ *AC* =

Given sec Î¸ = calculate all other trigonometric ratios.

SOLUTION:**Soln. :** In right Î”*ABC*, âˆ *B* = 90Â°

Let âˆ *A* = Î¸ and sec Î¸ =

â‡’ *AC* = 13*k* and *AB* = 12*k*

âˆ´ Using Pythagoras theorem, we get
*BC*^{2} = *AC*^{2} - *AB*^{2}

â‡’ *BC*^{2} = (13*k*)^{2} - (12*k*)^{2}

= (13*k* - 12*k*) (13*k* + 12*k*)

= *k*(25*k*) = 25*k*^{2} = (5*k*)^{2}

If âˆ *A* and âˆ *B* are acute angles such that cos *A* = cos *B* , then show that âˆ *A* = âˆ *B* .

**Soln. :** Let us consider a right Î”*ABC*, âˆ *C* = 90Â°

Now, cos *A* =

And

Since, cos *A* = cos *B*

Now, in the Î”*ABC*, two sides *AC* and *BC* are equal. [Proved above]

âˆ´ Their opposite angles are also equal.

âˆ´ âˆ *A* = âˆ *B*.

If cot Î¸ = evaluate:

(i) (ii) cot ^{2}Î¸

**Soln. :** In right Î”*ABC*, âˆ *B* = 90Â° and âˆ *A* = Î¸

.......(1) [Given]

But in right Î”*ABC*, cot Î¸ = .......(2)

From (1) and (2),

â‡’ *AB* = 7*k* and *BC* = 8*k*

âˆ´ *AC*^{2} = *AB*^{2} + *BC*^{2} = (7*k*)^{2} + (8*k*)^{2}

(Pythagoras theorem)

Now,

(i)

(ii)

If 3 cot *A* = 4, check whether

**Soln. :** In right angled Î”*ABC*, âˆ *B* = 90Â°

âˆ´ For âˆ *A*, we have:

Base = *AB* and perpendicular = *BC*. Also

Hypotenuse = *AC*

âˆ´ 3 cot *A* = 4

âˆ´ cot *A* = 4/3 ...(1)

But cot *A * ...(2)

âˆ´ From (1) and (2),

â‡’ *AB* = 4*k* and *BC* = 3*k*

âˆ´ Using Pythagoras theorem, we get
* AC*^{2} = *AB*^{2} + *BC*^{2}

â‡’ *AC*^{2} = (4*k*)^{2} + (3*k*)^{2}
^{ }

Now,

Now, to check the given equation,

L.H.S. =

........(i)

R.H.S. = cos^{2}*A* - sin^{2}*A*

.........(ii)

From (i) and (ii), we have L.H.S. = R.H.S.

In triangle *ABC* , right angled at *B* , if tan *A* =

find the value of :

(i) sin *A* cos *C* + cos *A* sin *C*

(ii) cos *A* cos *C* - sin *A* sin *C*

**Soln. :** In right Î”*ABC*, âˆ *B* = 90Â°

For âˆ *A*, we have

Base = *BC*, Perpendicular = *AB*,

Hypotenuse = *AC*

.......(1) (Given)

âˆ´ tan *A * ........(2)

From (1) and (2), we have

â‡’ *AB* = and *BC* = 1*k*

Now, using Pythagoras theorem, we get
*AC*^{2} = *AB*^{2} + *BC*^{2}
^{}
^{}

Now, ,

Again, for âˆ *C*, we have

Base = *BC*, Perpendicular = *AB*

and Hypotenuse = *AC*

,

(i)

(ii)

In Î”*PQR* , right-angled at *Q* , *PR* + *QR* = 25 cm and *PQ* = 5 cm. Determine the values of sin *P* , cos *P* and tan *P* .

**Soln. :** In right Î”*PQR*, *Q* = 90Â°

*PR* + *QR* = 25 cm and *PQ* = 5 cm

Let *QR* = *x* cm â‡’ *PR* = (25 - *x*)

âˆ´ By Pythagoras theorem, we have
*PR*^{2} = *QR*^{2} + *PQ*^{2}

â‡’ (25 - *x*)^{2} = *x*^{2} + 5^{2}

â‡’ 625 - 50*x* + *x*^{2} = *x*^{2} +25

â‡’ - 50*x* = - 600

â‡’ *i*.*e*., *RQ* = 12 cm

â‡’ *RP* = 25 - 12 = 13 cm

Now, , ,

State whether the following are true or false. Justify your answer.

(i) The value of tan *A* is always less than 1.

(ii) sec *A* = for some value of angle *A* .

(iii) cos *A* is the abbreviation used for the cosecant of angle *A* .

(iv) cot *A* is the product of cot and *A* .

(v) for some angle Î¸.

**Soln. :** (i) False [âˆµ *A* tangent of an angle is ratio of sides other than hypotenuse, which may be equal or unequal to each other.]

(ii) True [âˆµ cos *A* is always less than 1]

*i*.*e*., sec *A* will always be greater

than 1

(iii) False [âˆµ '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 sinÎ¸ cannot be greater than 1]

Evaluate the following:

(i) sin60Â° cos30Â° + sin 30Â° cos 60Â°

(ii) 2tan ^{2} 45Â° + cos ^{2} 30Â° - sin ^{2} 60Â°

(iii)

(iv)

(v)

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

(ii) 2tan^{2} 45Â° + cos^{2} 30Â° - sin^{2} 60Â°

(iii)

(iv)

and cot 45Â° = 1

(v)

Choose the correct option and justify your choice:

(i)

(A) sin 60Â° (B) cos 60Â°

(C) tan 60Â° (D) sin 30Â°

(ii)

(A) tan 90Â° (B) 1

(C) sin 45Â° (D) 0

(iii) sin 2 *A* = 2sin *A* is true when *A* =

(A) 0Â° (B) 30Â°

(C) 45Â° (D) 60Â°

(iv)

(A) cos 60Â° (B) sin 60Â° (C) tan 60Â° (D) sin 30Â°

**Soln. :** (i) (A)

(ii) (D) :

(iii) (A) : When *A* = 0, then

sin 2*A* = sin 2(0Â°) = sin 0Â° = 0,

2 sin *A *= 2 sin 0Â° = 2 Ã— 0 = 0
*i*.*e*., sin 2*A* = 2sin *A* for *A* = 0Â°

(iv) (C) :

If tan ( *A* + *B* ) = and tan ( *A* - *B* ) =

0Â° < *A* + *B* < 90Â°; *A* > *B* , find *A* and *B*.

**Soln. :** We have,

tan 60Â° , tan 30Â° = ......(1)

Also tan (*A* + *B*) and tan (*A* - *B*) (Given)

.......(2)

From (1) and (2), we get *A* + *B* = 60Â° and *A* - *B* = 30Â°

On adding *A* + *B* and *A* - *B*, we get 2*A* = 90Â°

â‡’ *A* = 45Â°

On subtracting *A* + *B* and *A* - *B*, we get 2*B* = 30Â°

â‡’ *B* = 15Â°

State whether the following are true or false. Justify your answer.

(i) sin ( *A* + *B* ) = sin *A* + sin *B* .

(ii) The value of sin Î¸ increases as Î¸ increases.

(iii) The value of cos Î¸ increases as Î¸ increases.

(iv) sin Î¸ = cos Î¸ for all values of Î¸ .

(v) cot *A* is not defined for *A* = 0Â°.

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

R.H.S.= sin 30Â° + sin 60Â°

âˆ´ L.H.S. â‰ R.H.S.

(ii) True : Since the values of sin Î¸ increases from 0 to 1 as Î¸ increases from 0 to 90Â°.

(iii) False : Since the value of cos Î¸ decreases from 1 to 0 as Î¸ increases from 0 to 90Â°.

(iv) False : Let us take Î¸ = 30Â°

â‡’ sin 30Â° â‰ cos 30Â°

(v) True : We have cot 0Â° = not defined

Evaluate:

(i) (ii)

(iii) (iv)

**Soln. :** (i)

â‡’ sin 18Â° = sin (90Â° - 72Â°) = cos 72Â°

[âˆµ sin (90Â° - *A*) = cos *A*]

âˆ´

(ii)

tan 26Â° = tan (90Â° - 64Â°) = cot 64Â° [âˆµ tan (90Â° - *A*) = cot *A*]

(iii) cos 48Â° - sin 42Â°

cos 48Â° = cos (90Â° - 42Â°) = sin 42Â° [âˆµ cos (90Â° - *A*) = sin *A*]

âˆ´ cos 48Â° - sin 42Â° = sin 42Â° - sin 42Â° = 0

(iv) cosec 31Â° - sec 59Â°

cosec 31Â° = cosec (90Â° - 59Â°) = sec 59Â° [âˆµ cosec (90Â° - *A*) = sec *A*]

âˆ´ cosec 31Â° - sec 59Â° = sec 59Â° - sec 59Â° = 0

Show that:

(i) tan 48Â° tan 23Â° tan 42Â° tan 67Â° = 1

(ii) cos 38Â° cos 52Â° - sin 38Â° sin 52Â° = 0

**Soln. :** (i) tan 48Â° tan 23Â° tan 42Â° tan 67Â° = 1

L.H.S. = tan 48Â° tan 23Â° tan 42Â° tan 67Â°

= tan (90Â° - 42Â°) tan 23Â° tan 42Â° tan (90Â° - 23Â°)

[Q tan (90Â° - *A*) = cot *A*]

= cot 42Â° tan 23Â° tan 42Â° cot 23Â°

âˆ´ L.H.S. = R.H.S.

(ii) cos 38Â° cos 52Â° - sin 38Â° sin 52Â° = 0

L.H.S. = cos 38Â° cos 52Â° - sin 38Â° sin 52Â°

= cos 38Â° cos (90Â° - 38Â°) - sin 38Â° sin (90Â° - 38Â°)

= cos 38Â° sin 38Â° - sin 38Â° cos 38Â°

[âˆµ sin (90Â° - *A*) = cos *A* and cos (90Â° - *A*)= sin *A*]

= 0 = R.H.S.

âˆ´ L.H.S. = R.H.S.

If tan 2 *A* = cot ( *A* - 18Â°), where 2 *A* is an acute angle, find the value of *A* .

**Soln. :** Since tan 2*A* = cot (*A* - 18Â°)

Also tan (2*A*) = cot (90Â° - 2*A*)

[âˆµ tan Î¸ = cot (90Â° - Î¸)]

â‡’ *A* - 18Â° = 90Â° - 2*A*

â‡’ *A* + 2*A* = 90Â° + 18Â°

â‡’ 3*A* = 108Â° â‡’

If tan *A* = cot *B* , prove that *A* + *B* = 90Â°

**Soln. :** tan *A* = cot *B* and cot *B* = tan (90Â° - *B*)

âˆ´ *A* = 90Â° - *B* â‡’ *A* + *B* = 90Â°

[âˆµ tan (90Â° - Î¸) = cot Î¸]

If sec 4 *A* = cosec ( *A* - 20Â°), where 4 *A* is an acute angle, find the value of *A*.

**Soln. :** âˆµ sec 4*A* = cosec (*A* - 20Â°)

sec (4*A*) = cosec (90Â° - 4*A*)

[âˆµ cosec (90Â° - Î¸) = sec Î¸]

âˆ´ *A* - 20Â° = 90Â° - 4*A*

â‡’ *A* + 4*A* = 90Â° + 20Â°

â‡’ 5*A* = 110Â° â‡’ *A* =

If *A* , *B* and *C* are interior angles of a triangle *ABC* , then show that

**Soln. :** Since, sum of the angles of Î”*ABC* is

*A* + *B* + *C* = 180Â°

âˆ´ *B* + *C* = 180Â° - *A*

Dividing both sides by 2, we get

[âˆµ sin (90Â° - Î¸) = cos Î¸]

Express sin 67Â° + cos 75Â° in terms of trigonometric ratios of angle between 0Â° and 45Â°.

SOLUTION:
**Soln. :** Since sin 67Â° = sin (90Â° - 23Â°) = cos 23Â°

[âˆµ sin (90Â° - Î¸ ) = cos Î¸]

Also, cos 75Â° = cos (90Â° - 15Â°) = sin 15Â°

[âˆµ cos (90Â° - Î¸) = sin Î¸]

âˆ´ sin 67Â° + cos 75Â° = cos 23Â° + sin 15Â°

Express the the trigonometric ratios sin *A* ,

sec *A* and tan *A* in terms of cot *A*.

**Soln. :** (a) sin *A *

(b)

(c)

Write all the other trigonometric ratios of âˆ *A* in terms of sec *A*.

**Soln. :** (i) sin *A* =

(ii)

(iii)

(iv)

(v)

Evaluate :

(i)

(ii)

**Soln. :** (i)

âˆµ sin 63Â° = sin (90Â° - 27Â°) = cos 27Â°

â‡’ sin^{2} 63Â° = cos^{2} 27Â°

cos^{2} 73Â° = cos^{2} (90Â° - 17Â°) = sin^{2} 17Â°

= 1

[âˆµ cos^{2}*A* + sin^{2}*A* = 1]

(ii) sin 25Â° cos 65Â° + cos 25Â° sin 65Â°

âˆµ sin 25Â° = sin (90Â° - 65Â°) = cos 65Â°

[âˆµ sin (90Â° - *A*) = cos *A*]

cos 25Â° = cos (90Â° - 65Â°) = sin 65Â°

[âˆµ cos (90Â° - *A*) = sin *A*]

âˆ´ sin 25Â° cos 65Â° + cos 25Â° sin 65Â°

= cos 65Â° cos 65Â° + sin 65Â° sin 65Â°

= (cos 65Â°)^{2} + (sin 65Â°)^{2}

= cos^{2} 65Â° + sin^{2} 65Â° = 1 [âˆµ cos^{2} *A* + sin^{2} *A* = 1]

Choose the correct option. Justify your choice.

(i) 9 sec ^{2} *A* - 9 tan ^{2} *A* = .................

(A) 1 (B) 9

(C) 8 (D) 0

(ii) (1 + tan Î¸ + sec Î¸ ) (1 + cot Î¸ - cosec Î¸ ) = ...................

(A) 0 (B) 1

(C) 2 (D) - 1

(iii) (sec *A* + tan *A* ) (1 - sin *A* ) = .......................

(A) sec *A* (B) sin *A*

(C) cosec *A* (D) cos *A*

(iv) .........................

(A) sec ^{2} *A* (B) - 1

(C) cot ^{2} *A* (D) tan ^{2} *A*

**Soln. :** (i) (B) : Since, 9 sec^{2} *A* - 9 tan^{2} *A*

= 9 (sec^{2} *A* - tan^{2} *A*) = 9 (1) = 9

[âˆµ tan^{2} *A* + 1 = sec^{2} *A* â‡’ sec^{2} *A* - tan^{2}*A* = 1]

(ii) (C) : Here, (1 + tan Î¸ + sec Î¸) (1 + cot Î¸ - cosec Î¸)

= (1 + tan Î¸ + sec Î¸)

= (1 + tan Î¸ + sec Î¸)

= 2

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

= cos *A*

(iv) (D) : Here,

= tan^{2} *A*

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosec Î¸ - cot Î¸ ) ^{2} =

(ii)

(iii)

[Hint : Write the expression in terms of sin Î¸ and cos Î¸ ]

(iv) [Hint : Simplify L.H.S. and R.H.S. separately]

(v) = cosec *A* + cot *A* , using the identity cosec ^{2} *A* = 1 + cot ^{2} *A* .

(vi)

(vii)

(viii) (sin *A* + cosec *A* ) ^{2} + (cos *A* + sec *A* ) ^{2}

= 7 + tan ^{2} *A* + cot ^{2} *A*

(ix) (cosec *A* - sin *A* ) (sec *A* - cos *A* )

[Hint : Simplify L.H.S. and R.H.S. separately]

(x)

**Soln. :** (i) L.H.S = (cosec Î¸ - cot Î¸)^{2}

[âˆµ sin^{2} Î¸ = 1 - cos^{2} Î¸]

[âˆµ 1 - cos^{2} Î¸ = (1 - cos Î¸) (1 + cos Î¸)]

Hence, L.H.S. = R.H.S.

(ii) L.H.S.

[âˆµ cos^{2} *A* + sin^{2} *A* = 1]

= R.H.S

Hence, L.H.S. = R.H.S.

(iii) L.H.S.

[âˆµ sin^{2} Î¸ + cos^{2} Î¸ = 1]

= cosec Î¸ sec Î¸ + 1 = 1 + sec Î¸Â·cosec Î¸

= R.H.S.

(iv)

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

= R.H.S. [âˆµ 1 - cos^{2} *A* = sin^{2} *A*]

(v) L.H.S.

[Dividing each term of num. and den. by sin *A*]

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

[âˆµ cot^{2} *A* - cosec^{2} *A* = - 1]

= cot *A* + cosec *A* = R.H.S.

(vi) L.H.S.

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

[âˆµ (1 - sin *A*) (1 + sin *A*)

= 1 - sin^{2} *A*]

[âˆµ 1 - sin^{2} *A* = cos^{2} *A*]

= sec *A* + tan *A* = R.H.S.

(vii)

(viii) L.H.S. = (sin *A* + cosec *A*)^{2} + (cos *A* + sec *A*)^{2}

= sin^{2} *A* + cosec^{2} *A* + 2 sin *A*Â·cosec *A* + cos^{2} *A*

+ sec^{2} *A* + 2 cos *A*Â·sec *A*

= (sin^{2} *A* + cos^{2} *A*) + cosec^{2} *A* + sec^{2} *A* + 2 + 2

[sin *A*Â·cosec *A* = 1 and sec *A*Â·cos *A* = 1]

= 1 + cosec^{2} *A* + sec^{2} *A* + 4

[âˆµ sin^{2} *A* + cos^{2} *A* = 1]

= 5 + (1 + cot^{2} *A*) + (1 + tan^{2} *A*)

[âˆµ cosec^{2} *A* = 1 + cot^{2} *A *and sec^{2} *A* = 1 + tan^{2} *A*]

= 7 + cot^{2} *A* + tan^{2} *A* = R.H.S

(ix) L.H.S. = (cosec *A* - sin *A*) (sec *A* - cos *A*)

[âˆµ 1 - sin^{2} *A* = cos^{2} *A*

and 1 - cos^{2} *A* = sin^{2} *A*]

= sin *A*Â·cos *A*
* *

[âˆµ 1 = sin^{2} *A* + cos^{2} *A*]

[Dividing num. and den. by sin *A* cos *A*]

(x)

= tan^{2} *A* = R.H.S. ...(1)

= (-tan *A*)^{2} = tan^{2}*A* = R.H.S. ...(2)

âˆ´ From (1) and (2), we get