Question - 11 : -
Provethat:
(i) (cos 11^o +sin 11^o) / (cos 11^o – sin 11^o) = tan 56^o
(ii) (cos9^o + sin 9^o) / (cos 9^o – sin 9^o)= tan 54^o
(iii)(cos 8^o – sin 8^o) / (cos 8^o + sin 8^o)= tan 37^o

Answer - 11 : -

(i) (cos 11^o + sin 11^o) / (cos11^o – sin 11^o) = tan 56^o

Let us consider LHS:

(cos 11^o + sin 11^o) / (cos11^o – sin 11^o)

Now let us divide the numerator and denominator by cos11^o we get,

(cos 11^o + sin 11^o) / (cos11^o – sin 11^o) = (1 + tan 11^o) / (1 – tan11^o)

= (1 + tan 11^o) / (1- 1×tan 11^o)

= (tan 45^o + tan 11^o) / (1– tan 45^o× tan 11^o)

We know that tan (A+B) = (tan A + tan B) / (1 – tan Atan B)

So,

(tan 45^o + tan 11^o) / (1 –tan 45^o× tan 11^o) = tan (45^o + 11^o)

= tan 56^o

= RHS

∴ LHS = RHS

Hence proved.

(ii) (cos 9^o + sin 9^o) / (cos 9^o –sin 9^o) = tan 54^o

Let us consider LHS:

(cos 9^o + sin 9^o) / (cos 9^o –sin 9^o)

Now let us divide the numerator and denominator by cos9^o we get,

(cos 9^o + sin 9^o) / (cos 9^o –sin 9^o) = (1 + tan 9^o) / (1 – tan 9^o)

= (1 + tan 9^o) / (1 – 1 × tan 9^o)

= (tan 45^o + tan 9^o) / (1 –tan 45^o × tan 9^o)

We know that tan (A+B) = (tan A + tan B) / (1 – tan Atan B)

So,

(tan 45^o + tan 9^o) / (1 –tan 45^o × tan 9^o) = tan (45^o + 9^o)

= tan 54^o

= RHS

∴ LHS = RHS

Hence proved.

(iii) (cos 8^o – sin 8^o) / (cos 8^o +sin 8^o) = tan 37^o

Let us consider LHS:

(cos 8^o – sin 8^o) / (cos 8^o +sin 8^o)

Now let us divide the numerator and denominator by cos8^o we get,

(cos 8^o – sin 8^o) / (cos 8^o +sin 8^o) = (1 – tan 8^o) / (1 + tan 8^o)

= (1 – tan 8^o) / (1 + 1×tan 8^o)

= (tan 45^o – tan 8^o) / (1 +tan 45^o ×tan 8^o)

We know that tan (A+B) = (tan A + tan B) / (1 – tan Atan B)

So,

(tan 45^o – tan 8^o) / (1 +tan 45^o ×tan 8^o) = tan (45^o – 8^o)

= tan 37^o

= RHS

∴ LHS = RHS

Hence proved.

Question - 12 : -
Provethat:
(i)
(ii)
(iii)

Answer - 12 : -

(i)

= sin 90^o

= 1

= RHS

∴ LHS = RHS

Hence proved.

(ii)

= sin 60^o

= √3/2

= RHS

∴ LHS = RHS

Hence proved.

(iii)

= sin 90^o

= 1

= RHS

∴ LHS = RHS

Hence proved.

Question - 13 : -
Provethat: (tan 69^o + tan 66^o) / (1 – tan 69^o tan66^o) = -1

Answer - 13 : -

Let us consider LHS:

(tan 69^o + tan 66^o) / (1 –tan 69^o tan 66^o)

We know that, tan (A + B) = (tan A + tan B) / (1 – tanA tan B)

Here, A = 69^o and B = 66^o

So,

(tan 69^o + tan 66^o) / (1 –tan 69^o tan 66^o) = tan (69 + 66)^o

= tan 135^o

= – tan 45^o

= – 1

= RHS

∴ LHS = RHS

Hence proved.

Question - 14 : -
(i) Iftan A = 5/6 and tan B = 1/11, prove that A + B = π/4
(ii) Iftan A = m/(m–1) and tan B = 1/(2m – 1), then prove that A – B = π/4

Answer - 14 : -

(i) If tan A = 5/6 and tan B = 1/11, prove that A + B =π/4

Given:

tan A = 5/6 and tan B = 1/11

We know that, tan (A + B) = (tan A + tan B) / (1 – tanA tan B)

= [(5/6) + (1/11)] / [1 – (5/6) × (1/11)]

= (55+6) / (66-5)

= 61/61

= 1

= tan 45^oor tan π/4

So, tan (A + B) = tan π/4

∴ (A + B) = π/4

Hence proved.

(ii) If tan A = m/(m–1) and tan B = 1/(2m – 1), thenprove that A – B = π/4

Given:

tan A = m/(m–1) and tan B = 1/(2m – 1)

We know that, tan (A – B) = (tan A – tan B) / (1 + tanA tan B)

= (2m² – m – m + 1) / (2m² –m – 2m + 1 + m)

= (2m² – 2m + 1) / (2m² –2m + 1)

= 1

= tan 45^o or tan π/4

So, tan (A – B) = tan π/4

∴ (A – B) = π/4

Hence proved.

Question - 15 : -
provethat:
(i) cos² π/4– sin² π/12 = √3/4
(ii) sin²(n+ 1) A – sin²nA = sin (2n + 1) A sin A

Answer - 15 : -

(i) cos² π/4 – sin² π/12 =√3/4

Let us consider LHS:

cos² π/4 – sin² π/12

We know that, cos²A – sin²B= cos (A + B) cos (A – B)

So,

cos² π/4 – sin² π/12 =cos (π/4 + π/12) cos (π/4 – π/12)

= cos 4π/12 cos 2π/12

= cos π/3 cos π/6

= 1/2 × √3/2

= √3/4

= RHS

∴ LHS = RHS

Hence proved.

(ii) sin²(n + 1) A – sin²nA =sin (2n + 1) A sin A

Let us consider LHS:

sin²(n + 1) A – sin²nA

We know that, sin²A – sin²B= sin (A + B) sin (A – B)

Here, A = (n + 1) A and B = nA

So,

sin²(n + 1) A – sin²n A =sin ((n + 1) A + nA) sin ((n + 1) A – nA)

= sin (nA +A + nA) sin (nA +A – nA)

= sin (2nA +A) sin (A)

= sin (2n + 1) A sin A

= RHS

∴ LHS = RHS

Hence proved.

Question - 16 : -
Provethat:
(i)
(ii)
(iii)
(iv) sin²B= sin²A + sin²(A-B) – 2sin A cos B sin (A –B)
(v) cos²A+ cos²B – 2 cos A cos B cos (A +B) = sin²(A +B)
(vi)

Answer - 16 : -

(i)

= tan A

= RHS

∴ LHS = RHS

Hence proved.

(ii)

Let us consider LHS:

= tan A – tan B + tan B – tan C + tan C – tan A

= 0

= RHS

∴ LHS = RHS

Hence proved.

(iii)

= cotB – cotA + cot C – cotB + cotA – cot C

= 0

= RHS

∴ LHS = RHS

Hence proved.

(iv) sin²B = sin²A + sin²(A-B)– 2sin A cos B sin (A – B)

Let us consider RHS:

sin²A + sin²(A -B) – 2 sinA cos B sin (A – B)

sin²A + sin (A -B) [sin (A –B) – 2 sin Acos B]

We know that, sin (A –B) = sin A cos B – cos A sin B

So,

sin²A + sin (A -B) [sin A cos B – cos A sinB – 2 sin A cos B]

sin²A + sin (A -B) [-sin A cos B – cos Asin B]

sin²A – sin (A -B) [sin A cos B + cos A sinB]

We know that, sin (A +B) = sin A cos B + cos A sin B

So,

sin²A – sin (A – B) sin (A + B)

sin²A – sin²A + sin²B

sin²B

= LHS

∴ LHS = RHS

Hence proved.

(v) cos²A + cos²B – 2cos A cos B cos (A + B) = sin²(A + B)

Let us consider LHS:

cos²A + cos²B – 2 cos A cos Bcos (A +B)

cos²A + 1 – sin²B – 2 cos A cosB cos (A +B)

1 + cos²A – sin²B – 2 cos Acos B cos (A +B)

We know that, cos²A – sin²B =cos (A +B) cos (A –B)

So,

1 + cos (A +B) cos (A –B) – 2 cos A cos B cos (A+B)

1 + cos (A +B) [cos (A –B) – 2 cos A cos B]

We know that, cos (A – B) = cos A cos B + sin A sin B.

So,

1 + cos (A +B) [cos A cos B + sin A sin B – 2 cosA cos B]

1 + cos (A +B) [-cos A cos B + sin A sin B]

1 – cos (A +B) [cos A cos B – sin A sin B]

We know that, cos (A +B) = cos A cos B – sin A sin B.

So,

1 – cos²(A + B)

sin²(A + B)

= RHS

∴ LHS = RHS

Hence proved.

(vi)

∴ LHS = RHS

Hence proved.

Question - 17 : -
Provethat:
(i) tan 8x –tan 6x – tan 2x = tan 8x tan 6x tan 2x
(ii) tanπ/12 + tan π/6 + tan π/12 tan π/6 = 1
(iii) tan36^o + tan 9^o + tan 36^o tan 9^o = 1
(iv) tan13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

Answer - 17 : -

(i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

Let us consider LHS:

tan 8x – tan 6x – tan 2x

tan 8x = tan(6x + 2x)

We know that, tan (A + B) = (tan A + tan B) / (1 – tanA tan B)

So,

tan 8x = (tan 6x + tan 2x) / (1 – tan 6x tan 2x)

By cross-multiplying we get,

tan 8x (1 – tan 6x tan 2x) = tan 6x + tan 2x

tan 8x – tan 8x tan 6x tan2x = tan 6x + tan 2x

Upon rearranging we get,

tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

= RHS

∴ LHS = RHS

Hence proved.

(ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1

We know,

π/12 = 15° and π/6 = 30°

So, we have 15° + 30° = 45°

Tan (15° + 30°) = tan 45°

We know that, tan (A + B) = (tan A + tan B) / (1 – tanA tan B)

So,

(tan 15^o + tan 30^o) / (1 –tan 15^o tan 30^o) = 1

tan 15° + tan 30° = 1 – tan 15° tan 30°

Upon rearranging we get,

tan15° + tan30° + tan15° tan30° = 1

Hence proved.

(iii) tan 36^o + tan 9^o + tan36^o tan 9^o = 1

We know 36° + 9° = 45°

So we have,

tan (36° + 9°) = tan 45°

We know that, tan (A + B) = (tan A + tan B) / (1 – tanA tan B)

So,

(tan 36^o + tan 9^o) / (1 –tan 36^o tan 9^o) = 1

tan 36° + tan 9° = 1 – tan 36° tan 9°

Upon rearranging we get,

tan 36° + tan 9° + tan 36° tan 9° = 1

Hence proved.

(iv) tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

Let us consider LHS:

tan 13x – tan 9x – tan 4x

tan 13x = tan (9x + 4x)

We know that, tan (A + B) = (tan A + tan B) / (1 – tanA tan B)

So,

tan 13x = (tan 9x + tan 4x) / (1 – tan 9x tan 4x)

By cross-multiplying we get,

tan 13x (1 – tan 9x tan 4x) = tan 9x + tan 4x

tan 13x – tan 13x tan 9x tan 4x = tan 9x + tan 4x

Upon rearranging we get,

tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

= RHS

∴ LHS = RHS

Hence proved.

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