## Trigonometric Ratios of Compound Angles

Subject: Optional Mathematics

#### Overview

Let A and B two angles. Then their sum A + B or the difference A - B is called a compound angle. The sum or difference of any two or more than two angles is called a compound angle.

##### Trigonometric Ratios of Compound Angles

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)

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

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

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

##### Things to remember
 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) =$\frac{cotA cotB + 1}{cotB - cotA}$
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##### How to solve problems based on trigonometric ratios of compound angles?

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

Here,

55° - 35° = 20°

Taking tan on both sides,

tan (55° - 35°) = tan 20°

or, $\frac {tan 55° - tan 35°}{1 + tan 55° tan35°}$ = tan 20°

or, tan 55° - tan 35° = tan 20° (1 + tan 55° tan35°)

or,tan 55° - tan 35° = tan 20° + tan 20° tan35° tan 55°

or, tan 55° - tan 35° = tan 20° + tan 20° tan (90° - 55°) tan 55°

or, tan 55° - tan 35° = tan 20° + tan 20° cot 55° tan 55°

or, tan 55° - tan 35° = tan 20° + tan 20°

∴ tan 55° - tan 35° = 2 tan 20°

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

L.H.S.

= cos (A + B + C)

= cos (A + B) cos C - sin (A + B) sin C

= (cos A cos B - sin A sin B) cos C - (sin A cos B + cos A sin B) sin C

multiply by $\frac {cos A cos B cos C}{cos A cos B cos C} = cos A cos B cos C (\(\frac {cos A cos B cos C}{cos A cos B cos C}$ - $\frac {sin A sin B sin C}{cos A cos B cos C}$ - $\frac {sin A cos Bsin C}{cos A cosB cos C}$ - $\frac {cos A sin B sin C}{cos A cos B cos C}$)

= cos A cos B cos C (1 - tan A tan B - tan A tan C - tan B tan C)

= cos A cos B cos C (1 - tan B tan C - tan C tan A - tan A tan B)

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

R.H.S.

= tan 62°

= $\frac {sin 62°}{cos 62°}$

= $\frac {sin (45° + 17°)}{cos (45° + 17°)}$

= $\frac {sin 45° cos 17° + cos 45° sin 17°}{cos 45° cos 17° - sin 45° sin 17°}$

= $\frac {\frac {1}{\sqrt 2} cos 17° + \frac {1}{\sqrt 2} sin 17°}{\frac {1}{\sqrt 2} cos 17°- \frac {1}{\sqrt 2} sin 17°}$

= $\frac {\frac {1}{\sqrt 2} (cos 17° + sin 17°)}{\frac {1}{\sqrt 2} (cos 17°- sin 17°)}$

= $\frac {cos 17° + sin 17°}{cos 17° - sin 17°}$

= L.H.S. Proved

R.H.S.

= tan 53°

= $\frac {sin 53°}{cos 53°}$

= $\frac {sin (45° + 8°)}{cos (45° + 8°)}$

= $\frac {sin 45° cos 8° + cos 45° sin 8°}{cos 45° cos 8° - sin 45° sin 8°}$

= $\frac {\frac {1}{\sqrt 2} cos 8° + \frac {1}{\sqrt 2} sin 8°}{\frac {1}{\sqrt 2} cos 8°- \frac {1}{\sqrt 2} sin 8°}$

= $\frac {\frac {1}{\sqrt 2} (cos 8° + sin 8°)}{\frac {1}{\sqrt 2} (cos 8°- sin 8°)}$

= $\frac {cos 8° + sin 8°}{cos 8° - sin 8°}$

= L.H.S. Proved

Here,

sin A = $\frac 35$ and cos B = $\frac 5{13}$

cos A = $\sqrt {1 - sin^2 A}$

= $\sqrt {1 - (\frac 35)^2}$

= $\sqrt {1 - \frac 9{25}}$

= $\sqrt {\frac {25 - 9}{25}}$

= $\sqrt {\frac {16}{25}}$

= $\frac 45$

sin B = $\sqrt {1 - cos^2 B}$

= $\sqrt {1 - (\frac 5{13})^2}$

= $\sqrt {1 - \frac {25}{169}}$

= $\sqrt {\frac {169 - 25}{169}}$

= $\sqrt {\frac {144}{169}}$

= $\frac {12}{13}$

Now,

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

= $\frac 45$⋅ $\frac 5{13}$ - $\frac 35$⋅ $\frac {12}{13}$

= $\frac {20}{65}$ - $\frac {36}{65}$

= $\frac {20 - 36}{65}$

= -$\frac {16}{65}$ Ans

Here,

sin A = $\frac 1{\sqrt {10}}$, sin B = $\frac 1{\sqrt 5}$

cos A = $\sqrt {1 - sin^2 A}$

= $\sqrt {1 - (\frac 1{\sqrt {10}})^2}$

= $\sqrt {1 - \frac 1{10}}$

= $\sqrt {\frac {10 - 1}{10}}$

= $\sqrt {\frac {9}{10}}$

= $\frac 3{\sqrt {10}}$

cos B = $\sqrt {1 - sin^2 B}$

= $\sqrt {1 - (\frac 1{\sqrt 5})^2}$

= $\sqrt {1 - \frac 15}$

= $\sqrt {\frac {5 - 1}{5}}$

= $\sqrt {\frac {4}{5}}$

= $\frac 2{\sqrt 5}$

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

or, sin (A + B) = $\frac 1{\sqrt {10}}$× $\frac 2{\sqrt 5}$ + $\frac 3{\sqrt {10}}$× $\frac 1{\sqrt 5}$

or, sin (A + B) = $\frac 2{\sqrt {50}}$ + $\frac 3{\sqrt {50}}$

or, sin (A + B) = $\frac {2 + 3}{\sqrt {50}}$

or, sin (A + B) = $\frac 5{\sqrt {50}}$

or, sin (A + B) = $\frac 5{5\sqrt 2}$

or, sin (A + B) = $\frac 1{\sqrt 2}$

or, sin (A + B) = sin $\frac {π^2}4$

∴ A + B =$\frac {π^2}4$ Proved

L.H.S.

= $\frac {cos 35° + sin 35°}{cos 35° - sin 35°}$

=$\frac {cos (45° - 10°) + sin (45° - 10°)}{cos (45° - 10°) - sin (45° - 10°)}$

= $\frac {cos 45° cos 10° + sin 45° sin 10° + sin 45° cos 10° - cos 45° sin 10°}{cos 45° cos 10° + sin 45° sin 10° - sin 45° cos 10° + cos 45° sin 10°}$

= $\frac {\frac 1{\sqrt 2} cos 10° + \frac 1{\sqrt 2} sin 10° + \frac 1{\sqrt 2} cos 10° - \frac 1{\sqrt 2} sin 10°}{\frac 1{\sqrt 2} cos 10° + \frac 1{\sqrt 2} sin 10° - \frac 1{\sqrt 2} cos 10° + \frac 1{\sqrt 2} sin 10°}$

=$\frac {\frac 2{\sqrt 2} cos 10°}{\frac 2{\sqrt 2} sin 10°}$

= $\frac {cos 10°}{sin 10°}$

= cot 10°

= R.H.S.Proved

L.H.S.

=cos (A + B)⋅cos (A - B)

= (cos A cos B - sin a sin B)(cos A cos B + sin A sin B)

= (cos A cos B)2 - (sin A sin B)2

= cos2A cos2B - sin2A sin2B

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

= cos2A - cos2A sin2B - sin2B + cos2A sin2B

= cos2A - sin2B (M.H.S.)

Again,

cos2A - sin2B

= (1 - sin2A) - (1 - cos2B)

= 1 - sin2A - 1 + cos2B

= cos2B - sin2A

∴ L.H.S. = M.H.S. = R.H.S. Proved

Given,

θ =α +β

sinθ = sin(α + β)............................(1)

And

$\frac {tanα}{tanβ}$ = $\frac xy$

$\frac {tanα - tanβ}{tanα + tanβ}$ = $\frac {x - y}{x + y}$.............................(2)

Now,

R.H.S.

= $\frac {x - y}{x + y}$ sinθ

=$\frac {tanα - tanβ}{tanα + tanβ}$ sinθ

=$\frac {\frac {sinα}{cosα} -\frac {sinβ}{cosβ}}{\frac {sinα}{cosα} + \frac {sinβ}{cosβ}}$ sinθ

= $\frac{\frac{sin\alpha.cos\beta - cos\alpha.sin\beta}{cos\alpha.cos\beta}}{\frac{sin\alpha.cos\beta + cos\alpha.sin\beta}{cos\alpha.cos\beta}}$. sin$\theta$

= $\frac {sin\alpha cos\beta - cos\alpha sin\beta}{sin\alpha cos\beta + cos\alpha sin\beta}$ . sin$\theta$

= $\frac {sin (\alpha - \beta)}{sin(\alpha + \beta)}$ . sin$\theta$

= $\frac {sin(\alpha - \beta)}{sin \theta}$. sin$\theta$

= sin (α - β)

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

L.H.S.

= sin (A + B + C)

= sin {A + (B + C)}

= sinA . cos(B + C) + cos A . cos (B + C)

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

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

= cosA cosB cosC[$\frac {sinA cosB cosC - sinA sinB sinC + cosA sinB cosC + cosA cosB sinC}{cosA cosB cosC}$]

= cosA cosB cosC [tanA - tanA tanB tanC + tanB + tanC]

= cosA cosB cosC [tanA + tanB + tanC- tanA tanB tanC ]

∴ L.H.S. = R.H.S. Proved