## Note on Trigonometric Ratios of Compound Angles

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Let, A and B be two angles. Then their sum A + B or the difference A - B is called a compound angle.

### Trigonometric ratios of A + B (Addition formula)

Let a revolving line start from OX and trace out an angle XOY = A and revolve further through an angle YOZ = B

∴ ∠XOZ = A + B

Let P be any point in OZ. Draw PM perpendicular to OX and PN perpendicular to OY From N draws NQ perpendicular to OX ad NR perpendicular to MP.

Here, ∠RPN = 90 - ∠PNR

= ∠RNO

= ∠NOQ

= A

Again, RMQN is a rectangle, so, MR = QN and RN = MQ

Now, sin(A + B) =$$\frac{MP}{OP}$$

= $$\frac{MR + RP}{OP}$$

= $$\frac{QN + RP}{OP}$$

= $$\frac{QN}{OP}$$ + $$\frac{RP}{OP}$$

= $$\frac{QN}{ON}$$ $$\frac{ON}{OP}$$ + $$\frac{RP}{NP}$$ $$\frac{NP}{OP}$$

cos(A + B) = $$\frac{OM}{OP}$$

= $$\frac{OQ - MQ}{OP}$$

= $$\frac{OQ - RN}{OP}$$

= $$\frac{OQ}{OP}$$ - $$\frac{RN}{OP}$$

= $$\frac{OQ}{ON}$$ $$\frac{ON}{OP}$$ - $$\frac{RN}{NP}$$ $$\frac{NP}{OP}$$

= cosA cosB - sinA sinB

Hence, sin formula of compound angle (A + B) is sin (A + B) = sinA cosB + cosA sinB and consine formula of compound angle (A + B) and cos(A + B) = cosA cosB - sinA sinB

### Trigonometric Ratios of A - B(Subtraction formula)

Let a revolving line start from OX and trace out an angle XOY = A and then revolve back through an angle YOZ = B

∴ ∠XOZ = A - B

Let P be any point in the Line OZ. Draw PM perpendicular to OX and PN perpendicular to OY.

From N Draw NQ perpendicular to OX and perpendicular to MP produced.

Here, ∠RPN = 900 - ∠PNR

= ∠PNY

= ∠XOY

= A

Again QMRN is a rectangle. So, QN = MR and QM = NR

Now sin(A-B) =$$\frac{PM}{OP}$$

= $$\frac{MR - PR}{OP}$$

=$$\frac{QN - PR}{OP}$$

= $$\frac{QN}{OP}$$ - $$\frac{PR}{OP}$$

=$$\frac{QN}{ON}$$ $$\frac{ON}{OP}$$ - $$\frac{PR}{NP}$$ $$\frac{NP}{OP}$$

= sinA cosB - cosA sinB

cos(A - B) = $$\frac{OM}{OP}$$

=$$\frac{OQ + QM}{OP}$$

=$$\frac{OQ + NR}{OP}$$

=$$\frac{OQ}{OP}$$ + $$\frac{NR}{OP}$$

=$$\frac{OQ}{ON}$$ $$\frac{ON}{OP}$$ - $$\frac{NR}{NP}$$ $$\frac{NP}{OP}$$

= cosA cosB + sinA sinB

Hence, sine formula of compound angle (A - B) is sin (A - B) = sinA cosB - cosA sinB and cosine formula of compound angle (A - B) is cos (A - B) = cosA cosB + sinA sinB

Alternative Method

Take a unit circle with centre at the origin. Let the circle intersect the X-axis at the point P. Then the coorinates of P are (1.0)

Let Q be another point on the circumference of the circle such that∠POQ = A. Then the coordinates of Q are (cosA, sinA).

Let R be another point on the cirumference of the circle such that ∠QOR = B

Then ∠POR = ∠POQ + ∠QOR = A + B

So coordinates of R are ( cos(A + B) , sin(A + B)).

Take a point S on the circumference such that ∠POS = -B.

Then coordinates of the points S are (cos(-B), sin(-B)) = (cosB, sin(-B))

Here, ∠SOQ = ∠SOP + ∠POQ = A + B and ∠POR = ∠POQ + ∠QOR = A + B

∴ ∠SOQ = ∠POR

So, arc QS = arc PR

∴ Chord QS = Chord PR.

Now by distance formula

PR2 = [cos(A + B)-1]2 + [sin(A + B) - 0]2

= cos2 (A + B) - 2cos(A + B) + 1 + sin2 (A + B)

= 2 - 2cos (A + B)

QS2 =(cosA - cosB)2 + [sinA - sin(-B)]2 = (cosA - cosB)2 + (sinA + sinB)2

= cos2A - 2cosA.cosB + cos2B + sin2A + 2sinA.sinB + sin2B

= 2 - 2cosA.cosB + 2sinA.sinB

Now, PR2 = QS2

or, 2 - 2cos(A + B) = 2 - 2cosA.cosB + 2sinA.sinB

or, cos(A + B) = cosA.cosB - sinA.sinB ........(i)

If the angle B is replaced by (-B), Then

Cos(A-B) = cosA.cos(-B) - sinA.sin(-B) = cosA.cosB + sinA.sinB .........(ii)

Again, cos[$$\frac{\pi}{2}$$ - (A+B)] = cos[($$\frac{\pi}{2}$$ - A) - B)]

or, sin(A + B) = cos ($$\frac {\pi}{2}$$ - A) cos B + sin ($$\frac{\pi}{2}$$ - A) sin B = sinA cosB + cosA sinB ....... (iii)

Similarly, cos[$$\frac{\pi}{2}$$ - (A+B)] = cos[($$\frac{\pi}{2}$$ - A) + B)]

or, sin(A - B) = cos ( $$\frac{\pi}{2}$$ - A) cosB - sin( $$\frac{\pi}{2}$$ - A) sinB = sinA cosB - cosA sinB .......... (iv)

### Tangent formula of compound angle (A + B)

tan (A + B) = $$\frac{sin(A + B)}{cos(A + B)}$$

= $$\frac{sinA\; cosB + cosA\; sinB}{cosA \;cosB - sinA \;sinB}$$

= $$\frac {\frac {sinA\; cosB}{cosA \;cosB} + \frac {cosA \;sinB}{cosA \;cosB}}{\frac {cosA \;cosB}{cosA\; cosB} - \frac {sinA\; sinB}{cosA \;cosB}}$$

= $$\frac{tan A + tan B}{1 - tanA\; tanB}$$

### Tangent formula of compound angle (A - B)

tan (A - B) = $$\frac{sin(A - B)}{cos(A - B)}$$

=$$\frac{sinA\; cosB - cosA \;sinB}{cosA \;cosB + sinA \;sinB}$$

=$$\frac{\frac{sinA\; cosB}{cosA\; cosB} - \frac{cosA\; sinB}{cosA \;cosB}}{\frac{cosA \;cosB}{cosA \;cosB} + \frac{sinA\; sinB}{cosA\; cosB}}$$

= $$\frac{tan A - tan B}{1 + tanA \;tanB}$$

### Cotangent formula of compound angle (A + B)

cot (A + B) = $$\frac{cos(A + B)}{sin(A + B)}$$

=$$\frac{cosA\; cosB - sinA \;sinB}{sinA\; cosB + cosA\; sinB}$$

=$$\frac{\frac{cosA \;cosB}{sinA\; sinB} - \frac{sinA\; sinB}{sinA\; sinB}}{\frac{sinA \;cosB}{sinA\; sinB} + \frac{cosA\; sinB}{sinA\; sinB}}$$

=$$\frac{cotA \;cotB - 1}{cotB + cotA}$$

### Cotangent formula of compound angle (A - B)

cot(A - B) = $$\frac{cos(A - B)}{sin(A - B)}$$

=$$\frac{cosA\; cosB + sinA\; sinB}{sinA \;cosB - cosA \;sinB}$$

=$$\frac{\frac{cosA \;cosB}{sinA\; sinB} + \frac{sinA \;sinB}{sinA \;sinB}}{\frac{sinA \;cosB}{sinA \;sinB} - \frac{cosA\; sinB}{sinA \;sinB}}$$

 Trigonometric Ratios of Compound Angles sin(A + B) = sinA cosB + cosA sinB sin(A - B) = sinA cosB - cosA sinB cos(A + B) = cosA cosB - sinA sinB cos(A - B) = cosA cosB + sinA sinB tan(A + B) = $$\frac{tan A + tan B}{1 - tanA\; tanB}$$ tan(A - B) =$$\frac{tanA - tanB}{1 + tanA \;tanB}$$ cot(A + B) =$$\frac{cotA\; cotB - 1}{cotB + cotA}$$ cot(A - B) =$$\frac{cotA \;cotB + 1}{cotB - cotA}$$
##### Some more results :

1. sin(A + B). sin(A - B) = cos2B - cos2A

Proof:

sin(A + B) .sin(A - B)

= (sinA cosB + cosA sinB) . (sinA cosB - cosA sinB)

= sin2A cos2B - cos2A sin2B

= (1 - cos2A) cos2B - cos2A(1 - cos2B)

= cos2B - cos2A cos2B - cos2A + cos2A cos2B

= cos2B - cos2A

2. sin(A + B). sin(A - B) = sin2A - sin2B

proof:

sin(A + B). sin(A - B)

= cos2B - cos2A

= 1 - sin2B - (1 - sin2A)

= sin2A - sin2B

3. cos(A + B). cos(A - B) = cos2A - sin2B

Proof :

cos (A + B). cos(A - B)

= (cosA cosB - sinA sinB) (cosA cosB + sinA sinB)

= cos2A cos2B - sin2A sin2B

= cos2A(1 - sin2B) - (1 - cos2A) sin2B

= cos2A - cos2A sin2B - sin2B + cos2A sin2B

= cos2A - cos2A sin2B - sin2B + cos2A sin2B

= cos2A - sin2A

4. cos(A + B) . cos(A - B) = cos2B - sin2A

Proof :

cos(A + B) . cos(A - B)

= cos2A - sin2B

= 1 - sin2A - (1 - cos2B)

= cos2B - sin2A

5. cot(A + B) .cot(A - B) =$$\frac{cot^{2} A. cot^{2} B - 1}{cot^{2} B - cot^{2} A}$$

Proof:

cot(A + B). cot(A - B)

= ( $$\frac{cotA. cotB - 1}{cotB + cotA}$$) ( $$\frac{cotA .cotB + 1}{cotB - cotA}$$)

=$$\frac{cot^{2}A . cot^{2}B}{cot^{2}B - cot^{2}A}$$

6. tan(A + B). tan(A - B) =$$\frac{tan^{2}A - tan^{2}B}{1 - tan^{2}A . tan^{2}B}$$

Proof :

tan(A + B). tan(A - B)

= ($$\frac{tanA + tanB}{1 - tanA.tanB}$$) ($$\frac{tanA - tanB}{1 + tanA tanB}$$)

= $$\frac{tan^{2}A - tan^{2}B}{1 - tan^{2}A tan^{2}B}$$

7. sin(A + B + C) = sinA cosB cosC + cosA sinB cosC + cosA cosB sinC - sinA sinB sinC

Proof :

sin(A + B + C)

= sin(A + B) cosC + cos(A + B) sinC

= (sinA cosB + cosA sinB) cosC + (cosA cosB - sinA sinB) sinC

= sinA cosB cosC + cosA sinB cosC + cosA cosB sinC - sinA sinB sinC

8. cos(A + B + C) = cosA.cosB.cosC - cosA.sinB.sinC - sinC.cosB.sinA - sinA.sinB.cosC

Proof:

cos(A + B + C)

= cos(A + B) cosC - sin(A + B) sinC

= (cosA cosB - sinA sinB) cosC - (sinA cosB + cosA sinB) sinC

= cosA.cosB.cosC - sinA sinB cosC - sinC.cosB.sinA - cosA.sinB.sinC

9. tan (A + B + C) =$$\frac{tanA + tanB + tanC - tanA tanB tanC}{1 - tanB tanC - tanC tanA - tanA tanB}$$

Proof:

tan(A + B + C)

= $$\frac{tan(A + B) + tanC}{1 - tan (A + B) tanC}$$

= $$\frac{\frac{tanA + tanB}{1 - tanA tanB} + tanC}{1 -(\frac{tanA + tanB}{1 - tanA tanB}) tanC}$$

=$$\frac{tanA + tanB + tanC - tanA tanB tanC}{1 - tanB tanC - tanC tanA - tanA tanB}$$

 Trigonometric Ratios of Compound Angles sin(A + B) = sinA cosB + cosA sinB sin(A - B) = sinA cosB - cosA sinB cos(A + B) = cosA cosB - sinA sinB cos(A - B) = cosA cosB + sinA sinB tan(A + B) = $$\frac{tan A + tan B}{1 - tanA tanB}$$ tan(A - B) =$$\frac{tanA - tanB}{1 + tanA tanB}$$ cot(A + B) =$$\frac{cotA cotB - 1}{cotB + cotA}$$ cot(A - B) =$$\frac{cotA cotB + 1}{cotB - cotA}$$
.

### Very Short Questions

Here,

cos75°

= cos (45° + 30°)

= cos45° ⋅ cos30° - sin45°⋅ sin30°

= $$\frac {1}{\sqrt 2}$$ ⋅$$\frac {\sqrt 3}{2}$$ - $$\frac 1{\sqrt 2}$$ ⋅$$\frac 12$$

= $$\frac {\sqrt 3}{2\sqrt 2}$$ - $$\frac 1{2\sqrt 2}$$

= $$\frac {\sqrt 3 - 1}{2\sqrt 2}$$ Ans

Here,

tan15°

= tan (60° - 45°)

= $$\frac {tan60° - tan45°}{1 + tan60° ⋅tan45°}$$

= $$\frac {\sqrt 3 - 1}{1 + \sqrt 3 ⋅ 1}$$

= $$\frac {\sqrt 3 - 1}{\sqrt 3 + 1}$$× $$\frac {\sqrt 3 - 1}{\sqrt 3 - 1}$$

= $$\frac {(\sqrt 3 - 1)^2}{{(\sqrt 3)^2}-{1^2}}$$

= $$\frac {3 - 2 {\sqrt 3 + 1}}{3 - 1}$$

= $$\frac {4 -2{\sqrt 3}}{2}$$

= $$\frac {2(2 - \sqrt 3)}{2}$$

= 2 - $$\sqrt 3$$Ans

sin 75° sin 15°

= sin (45° + 30°) sin (45° - 30°)

= (sin 45° cos 30° + cos45° sin30°) (sin 45° cos 30° - cos 45° sin30°)

= ($$\frac 1{\sqrt 2}$$⋅$$\frac {\sqrt 3}{2}$$ + $$\frac 1{\sqrt 2}$$⋅$$\frac {1}{2}$$)($$\frac 1{\sqrt 2}$$⋅$$\frac {\sqrt 3}{2}$$ -$$\frac 1{\sqrt 2}$$⋅$$\frac {1}{2}$$)

= ($$\frac {\sqrt 3}{2\sqrt 2}$$)2 -($$\frac {1}{2\sqrt 2}$$)2

= $$\frac 38$$ - $$\frac 18$$

= $$\frac {3 - 1}{8}$$

= $$\frac 28$$

= $$\frac 14$$ Ans

cos 105° cos 15°

= cos (60° + 45°) cos (60° - 45°)

= (cos 60° cos 45° - sin 60° sin 45°) (cos 60° cos 45° + sin 60° sin 45°)

= (cos 60° cos 45°)2 - (sin 60° sin 45°)2

= ($$\frac 12$$ × $$\frac {1}{\sqrt 2}$$)2 - ($$\frac {\sqrt 3}{2}$$ × $$\frac {1}{\sqrt 2}$$)2

= $$\frac 18$$ - $$\frac 38$$

= $$\frac {1 - 3}{8}$$

= $$\frac {-2}{8}$$

= $$\frac {-1}{4}$$ Ans

cos 105°

= cos (60° + 45°)

= cos 60° cos 45° - sin 60° sin 45°

= $$\frac 12$$⋅ $$\frac 1{\sqrt 2}$$ -$$\frac {\sqrt 3}2$$⋅ $$\frac 1{\sqrt 2}$$

= $$\frac {1 - \sqrt 3}{2\sqrt 2}$$ Ans

Here,

10° + 35° = 45°

Putting tan on both sides,

tan (10° + 35°) = tan 45°

or, $$\frac {tan 10° + tan 35°}{1 - tan 10° tan 35°}$$ = 1

or, tan 10° + tan 35° = 1 - tan 10° tan35°

or, 1 - tan 10° tan 35° = tan 10° + tan 35°

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

Here,

A + B = 45°

Putting tan on both;

tan (A + B) = tan 45°

or, $$\frac {tan A + tan B}{1 - tan A tan B}$$ = 1

or, tan A + tan B = 1 - tan A tan B

or, tan A + tan B + tan A tan B = 1

or, tan A + tan B + tan A tan B + 1 = 1 + 1

or, tan A + tan A tan B + 1 + tan B = 2

or, tan A (1 + tan B) + 1 (1 + tan B) = 2

or, (1 + tan B) (tan A + 1) = 2

∴ (1 + tan A) (1 + tan B) = 2

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

L.H.S.

= cot (A - B)

= $$\frac {cos (A - B)}{sin (A - B)}$$

= $$\frac {cos A cos B + sin A sin B}{sin A cos B - cos A sin B}$$

=$$\frac {\frac {cos A cos B}{sin A sin B}+ \frac{sin A sin B}{sin A sin B}}{\frac {sin A cos B}{sin A sin B}+ \frac{cos A sin B}{sin A sin B}}$$

= $$\frac {cot A cot B + 1}{cot B - cot A}$$

= R.H.S Proved

Here,

A + B = 45°

Taking tan on both sides:

tan (A + B) = tan 45°

or, $$\frac {tan A + tan B}{1 - tan A tan B}$$ = 1

or, tan A + tan B = 1 - tan A tan B

or, tan A + tan B + tan A tan B = 1

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

Here,

20° + 25° = 45°

Taking tan on both sides,

tan (20° + 25°) = tan 45°

or, $$\frac {tan 20° + tan 25°}{1 - tan 20° tan 25°}$$ = 1

or, tan 20° + tan 25° = 1 - tan 20° tan 25°

∴ 1 - tan 20° tan 25° = tan 20° + tan 25°

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

tan (α + β) = $$\frac {tanα + tanβ}{1 - tan α tan β}$$

or, tan (α + β) =$$\frac {\frac 56 + \frac 1{11}}{1 - \frac56 × \frac {1}{11}}$$

or, tan (α + β) =$$\cfrac {\frac {55 + 5}{66}}{\frac {66 - 5}{66}}$$

or, tan (α + β) =$$\cfrac {\frac {61}{66}}{\frac {61}{66}}$$

or, tan (α + β) = $$\frac {61}{66}$$× $$\frac {66}{61}$$

or, tan (α + β) = 1

or, tan (α + β) = tan $$\frac {π^c}{4}$$

∴ (α + β) = $$\frac {π^c}{4}$$ Proved

L.H.S.

= sin 105° + cos 105°

= sin (60° + 45°) + cos (60° + 45°)

= sin 60° cos 45° + cos 60° sin 45° + cos 60° cos 45° - sin 60° sin 45°

= $$\frac {\sqrt3}{2}$$⋅$$\frac 1{\sqrt 2}$$ + $$\frac 12$$⋅$$\frac 1{\sqrt 2}$$ + $$\frac 12$$⋅$$\frac 1{\sqrt 2}$$ - $$\frac {\sqrt 3}2$$⋅$$\frac 1{\sqrt 2}$$

= $$\frac {\sqrt 3}{2\sqrt 2}$$ + $$\frac 1{2\sqrt 2}$$ + $$\frac 1{2\sqrt 2}$$ - $$\frac {\sqrt 3}{2\sqrt 2}$$

= $$\frac {1 + 1}{2\sqrt 2}$$

= $$\frac 2{2\sqrt 2}$$

= $$\frac 1{\sqrt 2}$$

= R.H.S.

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

Given,

m sin(α + θ) = n sin (β + θ)

or, m (sin α cos θ + cos α sin θ) = n (sin β cos θ + cos β sin θ)

or, msin α cos θ + mcos α sin θ = nsin β cos θ + ncos β sin θ

or, msin α cos θ -nsin β cos θ =ncos β sin θ -mcos α sin θ

or,cos θ (msin α - nsin β) = sin θ (ncos β - mcos α)

or, $$\frac {cos θ}{sin θ}$$ = $$\frac {ncos β - mcos α}{msin α - nsin β}$$

∴ cot θ = $$\frac {ncos β - mcos α}{msin α - nsin β}$$ Proved

Here,

20° + 72° + 88° = 180°

20° + 72° = 180° - 88°

Putting tan on both,

tan (20° + 72°) = tan (180° - 88°)

or, $$\frac {tan 20° + tan 72°}{1 - tan 20° tan 72°}$$ = 0 - tan 88°

or, tan 20° + tan 72° = - tan 88° (1 - tan 20° tan 72°)

or, tan 20° + tan 72° = - tan 88° + tan 20° tan 72° tan 88°

or, tan 20° + tan 72° + tan 88° = tan 20° tan 72° tan 88°

Hence L.H.S. = R.H.S. Proved

Here,

L.H.S.

= sin (x + y) - sin ( x - y)

= sin x cos y + cos x sin y - (sin x cos y - cos x sin y)

=sin x cos y + cos x sin y - sin x cos y + cos x sin y

= 2 cos x sin y

= R.H.S.

Hence L.H.S. = R.H.S. Proved

Here,

A + B = $$\frac {π^c}{4}$$

Taking cot on both,

cot (A + B) = cot $$\frac {π^c}{4}$$

or, $$\frac {cot A cot B - 1}{cot B + cot A}$$ = 1

or, cot A cot B - 1 = cot B + cot A

or, cot A cot B - cot A - cot B = 1

or, cot A cot B - cot A - cot B + 1 = 1 + 1

or, cot A (cot B - 1) -1 (cot B - 1) = 2

or, (cot A - 1) (cot B - 1) = 2

∴ L.H.S. = R.H.S. Proved

0%
• ### Find the value without using a calculator .sin 15(^o)

(frac{sqrt{3}-1}{2 sqrt{2} })

(frac{sqrt{3}-3}{3 sqrt{3} })

(frac{sqrt{2}-1}{2 sqrt{3} })

90(^o)

• ### Find the value without using a calculator .Sin 75(^o)

(frac{sqrt{3}+3}{2sqrt{3}})

(frac{sqrt{3}+1}{3sqrt{1}})

(frac{sqrt{5}+2}{3sqrt{1}})

(frac{sqrt{3}+1}{2sqrt{2}})

• ### Find the value without using a calculator .Sin 105(^o)

(frac{sqrt{3+1}}{4sqrt{2}})

(frac{sqrt{3+1}}{2sqrt{2}})

(frac{sqrt{3+1}}{3sqrt{4}})

(frac{sqrt{2+1}}{3sqrt{1}})

• ### Find the value without using a calculator .Sin 135(^o)

(frac{1}{sqrt{2}})

(frac{2}{sqrt{1}})

(frac{2}{sqrt{-2}})

(frac{2}{sqrt{4}})

• ### Find the value without using calculator .Cos 105(^0)

(frac{1-sqrt{2}}{2sqrt{2}})

(frac{1-sqrt{3}}{2sqrt{2}})

(frac{1-sqrt{2}}{2sqrt{3}})

(frac{1-sqrt{1}}{1sqrt{1}})

• ### Find the value without using calculator . tan 15(^o)

2 - (sqrt{-3})

2 - (sqrt{3})

-2 + (sqrt{-3})

3 - (sqrt{-3})

• ### Evaluate without using a calculator. tan 165 (^o)

(sqrt{2}) - 3

1.005423

(sqrt{2}) - 1

(sqrt{3}) - 2

• ### tan 70(^o) = tan 20(^o) + 2 tan 50 (^o)

1.009 , 12

1 (frac{24}{24})

1 (frac{24}{25})

1 (frac{25}{24})

• ### tan 50 (^o) = tan 40 (^o) + 2 tan 10 (^o)

- (frac{112}{162}) , (frac{122}{129})

- (frac{119}{169}) , (frac{120}{169})

- (frac{119}{119}) , (frac{120}{119})

tan 60

• ### tan 65 (^0) - tan 25 (^0) = 2 tan 40 (^o)

1

- (frac{119}{169}) , (frac{120}{169})

11

0.554

3

-1.224

-1.009

(frac{1}{2})

• ### If sin A = (frac{3}{5}) & sin B = (frac{12}{13}) , find the value of cos (A - B).

(frac{54}{65})

(frac{66}{65})

(frac{56}{65})

(frac{55}{55})

• ### If cosα = (frac{4}{5}) and cosβ = (frac{12}{13}) then find the value of cos (α+β).

(frac{36}{35})

(frac{33}{35})

1.090

(frac{33}{65})

• ### If sin A = (frac{3}{5}) and cos B = (frac{15}{17}) then find the value of sin (A + B).

(frac{77}{85})

(frac{85}{85})

(frac{80}{84})

(frac{77}{77})

• ### If tan A = (frac{3}{4}) and tan B = (frac{5}{12}) then find the value of sin (A - B).

(frac{12}{60})

(frac{10}{60})

(frac{65}{15})

(frac{15}{65})

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4cosA.cos(60-A).cos(60 A) =cos3A

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what is the value of tan 3A

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Find the value of:Sin70-cos80 cos140