**Question 1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :**

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

**Solution:**

Using Pythagoras theorem for ΔABC

AC

^{2}= AB^{2}+ BC^{2}= (24 cm)

^{2}+ (7 cm)^{2}= (576 + 49) cm

^{2}= 625 cm

^{2}∴AC = 25 cm

(i)sin A = opp/hypsin A = 7/25

cos A = adj/hyp = 24/25

cos A = 24/25

(ii)sin C = opp/hypsin C = 24/25

cos C = adj/hyp

cos C = 7/25

**Question 2. In Fig. 8.13, find tan P – cot R.**

**Solution:**

Applying Pythagoras theorem for ΔPQR, we obtain

PR

^{2}= PQ^{2}+ QR^{2}(13 cm)

^{2}= (12 cm)^{2}+ QR^{2}169 cm

^{2}= 144 cm^{2}+ QR^{2}25 cm

^{2}= QR^{2}QR = 5 cm

tan P = opp/adj

tan P = 5/12

cot R = adj/opp

cot R = 5/12

tan P – cot R = 5/12 – 5/12 = 0

**Question 3. If sin A = 3/4, calculate cos A and tan A.**

**Solution:**

Using sin

^{2}A + cos^{2}A = 1(3/4)

^{2}+ cos^{2}A = 1cos

^{2}A = 1 – (3/4)^{2}= 1 – 9/16

cos A = 7^{1/2}/4tan A = sin A/cos A

tan A = (3/4)/(7

^{1/2}/4)

tan A = 3/7^{1/2}

**Question 4: Given 15 cot A = 8. Find sin A and sec A**

**Solution:**

Given, 15 cot A = 8

cot A = 8/15

tan A = 1/cot A

tan A = 15/8

Using, 1 + tan

^{2}A = sec^{2}A1 + (15/8)

^{2 }= sec^{2}A289/64 = sec

^{2}A

sec A = 17/8We know, cos

^{2}A = 1/sec^{2}Acos

^{2}A = 64/289sin

^{2}A = 1 – cos2Asin

^{2}A = 225/289

sin A = 15/17

**Question 5: Given sec **θ** = 13/12, calculate all other trigonometric ratios.**

**Solution.**

Using Pythagoras theorem,

sin θ = 5/13

cos θ = 12/13

tan θ = 5/12

cosec θ = 13/5

cot θ = 12/5

**Question 6: If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.**

**Solution:**

Let us consider a ΔABC in which CD ⊥ AB.

It is given that cos A = cos B

AD/AC = BD/BC … (1)

We need to prove ∠A = ∠B. To prove this, we need to extend AC to P such that BC = CP.

From equation (1), we obtain

AD/BD = AC/BC

AD/BD = AC/CP(BC = CP by construction)By using the converse of B.P.T (Basic Proportionality Theorem),

CD||BP

∠ACD = ∠CPB (Corresponding angles) … (3)

And, ∠BCD = ∠CBP (Alternate interior angles) … (4)

By construction, we have BC = CP.

∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (5)

From equations (3), (4), and (5), we obtain

∠ACD = ∠BCD … (6)

In ΔCAD and ΔCBD,

∠ACD = ∠BCD (Using equation (6))

∠CDA = ∠CDB (Both 90°)

Therefore, the remaining angles should be equal.

∴∠CAD = ∠CBD

⇒ ∠A = ∠B

**Question 7: If cot **θ** = 7/8, evaluate**

**(i) ((1 + sin**θ) * (1 – sinθ))/(1 + cosθ) * (1 – cosθ**)))**

**(ii) cot**^{2}θ

^{2}

**Solution:**

(i)Using (a + b) * (a – b) = a^{2}– b^{2}in numerator and denominatorWe get

(1 – sin

^{2}θ)/(1 – cos^{2}θ)Using sin

^{2}θ + cos^{2}θ = 1We get

cos

^{2}θ/sin^{2}θ = cot^{2}θNow

cot^{2}θ = (7/8)2 = 49/64

(ii)cot^{2}θ = (7/8)2 = 49/64

**Question 8. If 3 cot A = 4, Check whether (1 – tan**^{2}A)/(1 + tan^{2}A) = cos^{2}A – sin^{2}A

^{2}A)/(1 + tan

^{2}A) = cos

^{2}A – sin

^{2}A

**Solution.**

We know that, tanA = sinA / cosA ….(1)

Using (1) on L.H.S

= (1 – sin

^{2}A/cos^{2}A)/(1 + sin^{2}A/cos^{2}A)which on rearranging becomes

= (cos

^{2}A – sin^{2}A)/(cos^{2}A + sin^{2}A)Using the identity,

cos

^{2}A + sin^{2}A = 1LHS becomes

= (cos

^{2}A – sin^{2}A)This is equal to RHS.

LHS = RHS (for every value of cot A)

Hence, Proved.

**Question 9: In **Δ**ABC, right-angled at B. If tan A = 1/(3**^{1/2}), find the value of

^{1/2}), find the value of

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

**(ii) cos A cos C − sin A sin C**

**Solution:**

Using Pythagoras theorem

(AB)

^{2}+ (BC)^{2}= (AC)^{2}(31/2)

^{2}+ (1)^{2}= (AC)^{2}which gives

AC = 2 cm

Using formulas

sin A = 1/2

sin C = 3

^{1/2}/2cos A = 3

^{1/2}/2cos C = 1/2

Now, (i) sin A cos C + cos A sin C

Substituting the values

= (1/2) * (1/2) + (3

^{1/2}/2) * (3^{1/2}/2)= 1/4 + 3/4

= 1Now, (ii) cos A cos C − sin A sin C

Substituting the values

= (3

^{1/2}/2) * (1/2) – (1/2) * (3^{1/2}/2)= 3

^{1/2}/4 – 3^{1/2}/4

= 0

**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:**

Given that, PR + QR = 25

PQ = 5

Let PR be x cm.

Therefore, QR = 25 − x cm

Applying Pythagoras theorem in ΔPQR, we obtain

PR

^{2}= PQ^{2}+ QR^{2}x

^{2}= (5)^{2}+ (25 − x)^{2}x

^{2}= 25 + 625 + x^{2}− 50x50x = 650

x = 13

Therefore, PR = 13 cm

QR = (25 − 13) cm = 12 cm

Now,

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5

**Question 11. State whether the following are true or false. Justify your answer.**

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

**(ii) sec A = 12/5 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) sin **θ** = 4/3, for some angle **θ

**Solution:**

(i)Consider a ΔPQR, right-angled at Q as shown below.Here tan P = 12/5 which is surely greater than 1.

Therefore, the statement is false.

(ii)Consider ΔABC with AB = 5 cm, AC = 12 cm and BC = x cmUsing Pythagoras theorem in ΔABC

(AB)

^{2}+ (BC)2 = (AC)252 + x2 = 122

x = (144 – 25)1/2

x = (119)1/2

x = 10.9 cm

AB < BC < AC

So this triangle is valid,

Therefore, given statement is true.

(iii)Abbreviation used for cosecant A is cosec A. And cos A is the abbreviation used for cosine A.

Hence, the given statement is false.

(iv)cot A is not the product of cot and A. It is the cotangent of ∠A.

Hence, the given statement is false.

(v)sin θ = 4/3In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible.

Hence, the given statement is false

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