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Given as

cos A = -12/13 and cot B = 24/7

As we know that, A lies in second quadrant, B in the third quadrant.

Since, in the second quadrant sine function is positive.

Since, in the third quadrant, both sine and cosine functions are negative.

On using the formulas,

sin A = √(1 – cos² A), sin B = – 1/√(1 + cot² B) and cos B = -√(1 – sin² B),

Therefore let us find the value of sin A and sin B

sin A = √(1 – cos² A)

= √(1 – (-12/13)²)

= √(1 – 144/169)

= √((169 - 144)/169)

= √(25/169)

= 5/13

sin B = – 1/√(1 + cot² B)

= – 1/√(1 + (24/7)²)

= – 1/√(1 + 576/49)

= -1/√((49 + 576)/49)

= -1/√(625/49)

= -1/(25/7)

= -7/25

cos B = -√(1 – sin² B)

= -√(1 - (-7/25)²)

= -√(1 - (49/625))

= -√((625 - 49)/625)

= -√(576/625)

= -24/25

Therefore, now let us find

(i) sin (A + B)

As we know that sin (A + B) = sin A cos B + cos A sin B

Therefore,

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

= 5/13 × (-24/25) + (-12/13) × (-7/25)

= -120/325 + 84/325

= -36/325

(ii) cos (A + B)

As we know that cos (A + B) = cos A cos B – sin A sin B

Therefore,

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

= -12/13 × (-24/25) – (5/13) × (-7/25)

= 288/325 + 35/325

= 323/325

(iii) tan (A + B)

As we know that tan (A + B) = sin (A + B)/cos (A + B)

= (-36/325)/(323/325)

= -36/323

Correct choice is B.

Since tan θ = -4/3 is negative, θ lies either in second quadrant or in fourth quadrant. Thus sin θ =4/5 if θ lies in the second quadrant or sin θ = -4/5, if θ lies in the fourth quadrant

Also, we know that cos function attains its max value 1 at 0 radians.

so,  x = y = 0

• x+y = 0

Hence, the maximum value of cos x + cos y + cos (x+y) = 1 + (1)+1=3

Thus, the maximum value of cos x + cos y + cos (x+y) = 3

We know that the cos function attains its minimum value as -1 at π radians.

• So, x = y = π
• x + y = 2π Now, we can write as;
• cos(x + y) = cos 2π = 1
• This is the min value of cos(x + y).

Thus, the minimum value of cos x + cos y + cos (x + y) = -1 + (-1 ) +1 = -1

Also, we know that cos function attains its max value 1 at 0 radians.

so,  x = y = 0

• x+y = 0

Hence, the maximum value of cos x + cos y + cos (x + y) = 1 + (1 )+1 = 3

Thus, the maximum value of cos x + cos y + cos (x + y) = 3 and minimum value = -1

RHS = (b + c)² − 4bc cos²\(\frac{A}{2}\).

= b² + c² + 2bc − 4bc \((\frac{cos\,A + 1}{2})\)

= b² + c² + 2bc – 2bc (cos A + 1)

= b² + c² − 2bc .\(\frac{b^2+c^2-a^2}{2bc}\)

= b² + c² – (b² + c² – a²)

= b² + c² – b² – c² + a²

= a² = LHS

∴ LHS = RHS

LHS = \(\frac{b^2-c^2}{a^2}\)

= \(\frac{k^2\,sin^2\,B -k^2sin^2\,C}{k^2\,sin^2A}\)

= \(\frac{sin^2\,B -sin^2\,C}{sin^2A}\)

= \(\frac{sin\,(B+C) -sin(B-C)}{sin^2[180°-(B+C)]}\)

= \(\frac{sin\,(B+C) -sin(B-C)}{sin^2(B+C)}\)

= \(\frac{sin(B-C)}{sin(B+C)}\) = RHS

∴ LHS = RHS

cos A = \(\frac{b^2+c^2-a^2}{2bc}\)

\(=\frac{5^2+7^2-3^2}{2.5.7}\)

\(=\frac{25+49-9}{70}\)

\(=\frac{65}{70}\)

 \(=\frac{13}{14}\)

cos B = \(\frac{a^2+c^2-b^2}{2bc}\)

 \(=\frac{3^2+7^2-5^2}{2.3.7}\)

\(=\frac{9+49-25}{42}\)

 \(=\frac{33}{42}\)

\(=\frac{11}{14}\)

cos C = \(\frac{a^2+b^2-c^2}{2ab}\)

\(=\frac{3^2+5^2-7^2}{2.3.5}\)

\(=\frac{9+25-49}{30}\)

\(=-\frac{15}{30}\)

\(=-\frac{1}{2}\)

Correct option is: (a) a² – b² 

\(\frac{sin\,A}{2sin\,{B}}\) = cos C

Sin A = 2 sin B cos C

ka= 2kb \(\frac{a^2+b^2-c^2}{2ab}\)

a = \(\frac{a^2+b^2-c^2}{2ab}\) 

a² = a² + b² – c²

⇒ b² – c² = 0

⇒ b² = c²

⇒ b = c

Hence, the triangle is isosceles.

\(\frac{a}{sin\,A}=\frac{b}{sin\,B}=\frac{c}{sin\,C}=k\)

⇒ k sin A, b = k sin B and c = k sin C

LHS = \(\frac{a+b}{c}=\frac{k\,sin\,A+k\,sin\,B}{k\,sin\,C}\)

\(=\frac{k(sin\,A+sin\,B)}{k\,sin\,C}\)

\(=\frac{sin\,A+sin\,B}{sin\,C}\) 

\(=\frac{2sin\,\frac{A+B}{2}\,cos\frac{A-B}{2}}{2\,sin\frac{C}{2}cos\frac{C}{2}}\)

\(=\frac{sin\,(90°-\frac{C}{2})cos\frac{A-B}{2}}{\,sin\frac{C}{2}cos\frac{C}{2}}\)

[∴ \(\frac{A+B}{2}\)= 90° − \(\frac{C}{2}\)]

\(=\frac{cos\frac{C}{2}cos\frac{A-B}{2}}{\,sin\frac{C}{2}cos\frac{C}{2}}\)

\(=\frac{cos\frac{A-B}{2}}{\,sin\frac{C}{2}}\) = RHS

∴ LHS = RHS

x² - 2x/(sin2A) + 1

\(\frac{sin\,A}{a}=\frac{sin\,B}{b}=\frac{sin\,C}{c}\)

\(\frac{sin\,A}{18}=\frac{sin\,B}{24}=\frac{sin\,90°}{30}\) 

\(\frac{sin\,A}{18}=\frac{sin\,B}{24}=\frac{1}{30}\)

\(sin\,A=\frac{18}{30}=\frac{3}{5}\)

\(sin\,B=\frac{24}{30}=\frac{4}{5}\)

LHS = \(a\,cos\frac{B-C}{2}\)

\(=k\,sin\,A\,cos\frac{B-C}{2}\)

\(=k\,sin\,(B+C)\,cos\frac{B-C}{2}\)

[ ∴ A + B +C = 180° ∴ sin A = sin [180° − (B + C) ]

\(=\frac{cos\frac{C}{2}cos\frac{A-B}{2}}{sin\frac{C}{2}cos\frac{C}{2}}\)

\(=\frac{cos\frac{A-B}{2}}{sin\frac{C}{2}}\) = RHS

∴ LHS = RHS

We know that,

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

∴ tan(A + B) = \(\frac{\frac{a}{a+1}+\frac{1}{2a+1}}{1-(\frac{a}{a+1})(\frac{1}{2a+1})}\)

⇒ tan(A + B) = \(\frac{a(2a+1)+(a+1)}{(a+1)(2a+1)-a}\)

⇒ tan(A + B) = \(\frac{2a^2+2a+1}{2a^2+2a+1}\)

⇒ tan(A + B) = 1

⇒ tan(A + B) = tan \((\frac{x}{4})\)

⇒ A + B = \(\frac{x}{4}\)

LHS = a [cos C – cos B]

= \(a[2\,sin{\frac{B+C}{2}sin\frac{B-C}{2}}]\)

\(= a[2\,cos{\frac{A}{2}sin\frac{B-C}{2}}]\)

\(= 2k\,sin\,A[cos{\frac{A}{2}sin\frac{B-C}{2}}]\)

\(= 2k\,2\,sin\,\frac{A}{2}\,cos\frac{A}{2}[cos{\frac{A}{2}sin\frac{B-C}{2}}]\)

\(= 2k\,cos^2\frac{A}{2}[2\,sin{\frac{A}{2}sin\frac{B-C}{2}}]\)

\(= 2k\,cos^2\frac{A}{2}k[sin\{90°-(\frac{B+C}{2})\}sin\frac{B-C}{2}]\)

 \(= 2\,cos^2\frac{A}{2}k[cos(\frac{B+C}{2})sin(\frac{B-C}{2})]\)

\(= 2\,cos^2\frac{A}{2}k[sin\,B\,-sin\,C]\)

\(= 2\,cos^2\frac{A}{2}[k\,sin\,B\,-k\,sin\,C]\)

\(= 2\,cos^2\frac{A}{2}[b \,-\,c]\) = RHS

∴ LHS = RHS

cos(54° + A) = sin(36° – A) 

cos(54° - A) = sin(36° + A) 

cos (36° – A) cos (36° +A)+ sin(36° - A)sin(36° + A) = cos(2A)

Correct answer is A.

Given expression is cos⁴ x + sin⁴ x – 6 cos² x sin² x

= [(cos²x)² + (sin²x)² - 2 cos²x sin²x] - 4 cos²x sin²x

[using the formula a² + b² = (a + b)² - 2ab]

= (cos²x - sin²x)2 - 4 cos²x sin²x

[using the formula cos 2x = cos² x – sin² x]

= (cos²x)² – (2 sinx cosx )²

[ using the formula sin 2x = 2 sin x cos x ]

= (cos 2x)² – (sin 2x)²

[ using the formula cos 2x = cos² x – sin² x ]

= cos 4x

Therefore cos⁴ x + sin⁴ x – 6 cos² x sin² x = cos 4x

cos² A – sin² B 

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

= cos \((\frac{π}3)\) cos (2x)

= \(\frac{1}2\) cos(2x)

A + B = π - C

\(\frac{tanA+tanB}{1-tanAtanB}\) = - tan c

tan A + tan B + tan C = tanA tanB tanC

cot A cot B cot C = \(\frac{1}6\)

3sin(x) + 4cos(x) = 5

\(\frac{3}5\) sin x + \(\frac{4}5\) cos x = 1

cos(37°- x) = cos 0° (∵ x = 37°)

4sin(x) - 3cos(x) = k

\(4\times\frac{3}5-3\times\frac{4}5\) = 0

A + B = C 

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

tan C – tan A – tan B = tan A tan B tan C.

Correct answer is A.

Given expression

cos (36° – A) cos (36° + A) + cos(54° – A) cos (54° + A)

In the above expression angle cos(54° + A) = sin[90° - (54° + A)]

And cos(54° - A) = sin[90° - (54° + A)]

[using cos θ = sin (90° - θ)]

Now substituting the same in the expression

= cos (36° – A) cos (36° + A) + sin[90°- (54°– A)] sin[90°- (54° + A)]

= cos (36° – A) cos (36° + A) + sin (36° + A) sin (36° - A)

= cos (36° + A) cos (36° – A) + sin (36° + A) sin (36° - A)

[using cos (A - B) = cos A cos B + sin A sin B]

= cos [(36° + A) – (36° - A)]

= cos (2A)

put a = 0 

tan(A) = 0 

tan(B) =1

tan A + B = \(\frac{tanA+tanB}{1-tanAtanB}\)

tan A + B =  \(\frac{0+1}{1-0}\)

tan A + B = 1

tan A + B = tan^-11

tan A + B = \(\frac{π}4\)

 a = b cos \(\frac{2π}3\) = c cos\(\frac{4π}3\) = k

a = k

b—2k 

c=-2k 

ab + bc + ca.= 0

L.H.S

= 2sin75°sin15°

= 2sin(45° + 30°)sin(45° - 30°)

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

= cos(-60°) - cos90°

= cos60° - 0

= \(\frac{1}{2}\)

(i) Let us consider the LHS

cos² π/4 – sin² π/12

As we know that, cos²A – sin² B = cos (A + B) cos (A – B)

Therefore,

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

Thus proved.

(ii) Let us consider the LHS

sin²(n + 1) A – sin²nA

As we know that, sin²A – sin² B = sin(A + B) sin(A – B)

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

Therefore,

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

Thus proved.

We have 36° + 9° = 45° 

⇒ tan(36° + 9°) = tan 45°

We know that tan(A + B) = \(\frac{tanA+tanB}{1-tanAtanB}\)

⇒ \(\frac{tan36°+tan9°}{1-tan36°tan9°}\) = 1

⇒ tan 36° + tan 9° = 1 – tan 36° tan 9° 

∴ tan 36° + tan 9° + tan 36° tan 9° = 1 

Hence proved.

⇒ π/12 = 15° And π/6 = 30° 

We have 15° + 30° = 45° 

⇒ tan(15° + 30°) = tan 45° 

We know that tan(A + B) = \(\frac{tanA+tanB}{1-tanAtanB}\)

⇒ \(\frac{tan15°+tan30°}{1-tan15°tan30°}\) = 1

⇒ tan15° + tan30° = 1 – tan15° tan30° 

∴ tan15° + tan30° + tan15° tan30° = 1 

Hence, proved.

sin x+ cosec x = 2

⇒ sin x + \(\cfrac1{sin\,\text x}\)

⇒ sin²x +1 = 2sin x

⇒ sin²x - 2sin x+1 = 0

⇒ (sin x - 1)² = 0

⇒ sin x = 1

As sin x = 1

sinⁿx = 1

∴ sinⁿ x + cosecⁿx

⇒ sinⁿ x + \(\cfrac1{sin^n\,\text x}\)= 1 + 1

⇒ 2.

We have 8x = 6x + 2x 

⇒ tan 8x = tan(6x + 2x) 

We know that tan(A + B) = \(\frac{tanA+tanB}{1-tanAtanB}\)

⇒ tan 8x = \(\frac{tan\,6x+tan\,2x}{1-tan\,6x\,tan\,2x}\)

⇒ tan8x (1 – tan6x tan2x) = tan6x + tan2x 

⇒ tan8x – tan8x tan 6x tan2x = tan6x + tan2x 

∴ tan8x – tan6x – tan2x = tan8x tan6x tan2x 

Hence, proved.

Given sin x = cos²x

To find the value of cos² x (1 + cos² x).

⇒ cos² x (1 + cos² x).

⇒ cos² x + cos⁴ x.

As cos²x = 1- sin²x the above equation becomes

⇒ 1- sin²x + sin²x

⇒ 1

(i) Let us consider the LHS

tan 8x – tan 6x – tan 2x

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

As we know that, tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

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

On 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

Now, upon rearranging we get,

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

= RHS

∴ LHS = RHS

Thus proved.

(ii) As we know,

π/12 = 15° and π/6 = 30°

Therefore, we have 15° + 30° = 45°

tan(15° + 30°) = tan 45°

As we know that, tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

(tan 15° + tan 30°)/(1 – tan 15° tan 30°) = 1

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

Now, upon rearranging we get,

tan15° + tan30° + tan15° tan30° = 1

Thus proved.

(iii) As we know 36° + 9° = 45°

Therefore we have,

tan (36° + 9°) = tan 45°

As we know that, tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

(tan 36° + tan 9°)/(1 – tan 36° tan 9°) = 1

tan 36° + tan 9° = 1 – tan 36° tan 9°

Now, upon rearranging we get,

tan 36° + tan 9° + tan 36° tan 9° = 1

Thus proved.

(iv) Let us consider the LHS

tan 13x – tan 9x – tan 4x

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

As we know that, tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

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

On 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

Now, upon rearranging we get,

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

= RHS

∴ LHS = RHS

Thus proved.

Value of cos(x) varies from -1 to 1 for all R and sin(x) is increasing in [-π/2,π/2]

∴ sin(cos x) has max value of sin1.

As we know that the maximum value of A cos α + B sin α + C is C + √(A² + B²),

And here the minimum value is C – √(a² + B²).

(i) Given as f(x) = 12 sin x – 5 cos x

Here, A = -5, B = 12 and C = 0

–√((-5)² + 12²) ≤ 12 sin x – 5 cos x ≤ √((-5)² + 12²)

–√(25 + 144) ≤ 12 sin x – 5 cos x ≤ √(25 + 144)

–√169 ≤ 12 sin x – 5 cos x ≤ √169

-13 ≤ 12 sin x – 5 cos x ≤ 13

Thus, the maximum and minimum values of f(x) are 13 and -13 respectively.

(ii) 12 cos x + 5 sin x + 4

Given as f(x) = 12 cos x + 5 sin x + 4

Here, A = 12, B = 5 and C = 4

4 – √(12² + 5²) ≤ 12 cos x + 5 sin x + 4 ≤ 4 + √(12² + 5²)

4 – √(144 + 25) ≤ 12 cos x + 5 sin x + 4 ≤ 4 + √(144 + 25)

4 – √169 ≤ 12 cos x + 5 sin x + 4 ≤ 4 + √169

-9 ≤ 12 cos x + 5 sin x + 4 ≤ 17

Thus, the maximum and minimum values of f(x) are -9 and 17 respectively.

(iii) 5 cos x + 3 sin (π/6 – x) + 4 

Given as f(x) = 5 cos x + 3 sin (π/6 – x) + 4 

As we know that, sin (A – B) = sin A cos B – cos A sin B

f(x) = 5 cos x + 3 sin (π/6 – x) + 4 

= 5 cos x + 3 (sin π/6 cos x – cos π/6 sin x) + 4

= 5 cos x + 3/2 cos x – 3√3/2 sin x + 4

= 13/2 cos x – 3√3/2 sin x + 4

Therefore, here A = 13/2, B = – 3√3/2, C = 4

4 – √[(13/2)² + (-3√3/2)²] ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + √[(13/2)² + (-3√3/2)²]

4 – √[(169/4) + (27/4)] ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + √[(169/4) + (27/4)]

4 – 7 ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + 7

-3 ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 11

Thus, the maximum and minimum values of f(x) are -3 and 11 respectively.

(iv) sin x – cos x + 1

Given as f(x) = sin x – cos x + 1

Therefore, here A = -1, B = 1 And c = 1

1 – √[(-1)² + 1²] ≤ sin x – cos x + 1 ≤ 1 + √[(-1)² + 1²]

1 – √(1 + 1) ≤ sin x – cos x + 1 ≤ 1 + √(1 + 1)

1 – √2 ≤ sin x – cos x + 1 ≤ 1 + √2

Thus, the maximum and minimum values of f(x) are 1 – √2 and 1 + √2 respectively.

sin(x) has maximum value at x = π/2 and its minimum at

x = -π/2 which are 1 and -1 respectively.

As 1 < π/2;

so, the argument of the outer sin always lies within the interva

 [-π/2, π/2]

So the maximum and minimum of the given function are 

in 1 and - sin 1.

Let cos x = t

Range of t =(-1,1)

∴ Maximum and Minimum value of cos x is 1 and -1 respectively.

Now,

cos(-x) = cos x

∴ Range of cos(cos x) = [cos(1),cos(0)]

⇒ cos(cos x) = [cos1,0]

the maximum value of (a cosx + bsinx) = \(\sqrt{a^2+b^2}\)

So maximum value = \(\sqrt{9+16}\)

= 5 

the minimum value of (a cosx + b sinx) = - \(\sqrt{a^2+b^2}\)

minimum value = - 5

f(x) = 12sin x - 9(sinx)²

f(x) = - (3sin x - 2)² + 4

- 1 ≤ sin x ≤ 1

f(x)ϵ-(3[-1,1]-2)²+4 

f(x)ϵ-([-3,3]-2)²+4 

f(x)ϵ-([-5,1])²+4 

f(x)ϵ-[0,25]+4 

f(x)ϵ[-25,0]+4 

f(x)ϵ[-21,4]

f(x) =12sin(x) - 9sin 2(x) 

f’(x) = 12cos(x) -18sin(x)cos(x) 

f’(x) = 0 for the maximum value of f(x) 

12cos(x) -18sin(x)cos(x) = 0

⇒ sinx = \(\frac{2}3\)

tan (x - y) = \(\frac{tanx-tany}{1+tanxtany}\)

tan[(A + B) - (A - B)] = \(\frac{tan(A+B)-tan(A-B)}{1+tan(A-B)tan(A+B)}\)

tan (2B) = \(\frac{p-q}{1+pq}\)

We know the range of sin x is

-1 ≤ sin x  ≤ 1

∴ 0 ≤ sin 14x  ≤ 1

We know the range of cos x is

-1 ≤ cos x  ≤ 1

∴ 0 ≤ cos 20x ≤ 1

0 < sin¹⁴x + cos²⁰x ≤ 2

which means that the value of x lies in the interval [0,2]

But there's a problem, when sine is 0 cosine is 1, they might even be 0 and -1 at particular points (not in this case, since they are even powers), so the minimum we would get should be more than 0. Hence the value of x lies in (0,1]

Given as 

sin x = 12/13 and x lies in the second quadrant.

As we know, in second quadrant, sin x and cosec x are positive and all other ratios are negative.

On using the formulas,

cos x = √(1 - sin² x)

= – √(1 - (12/13)²)

= – √(1 - (144/169))

= – √(169 - 144)/169

= -√(25/169)

= – 5/13

As we know,

tan x = sin x/cos x

sec x = 1/cos x

Then,

tan x = (12/13)/(-5/13)

= -12/5

sec x = 1/(-5/13)

= -13/5

sec x + tan x = -13/5 + (-12/5)

= (-13 - 12)/5

= -25/5

= -5

Thus, sec x + tan x = -5

Given as

sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2 

As we know that, x is in second quadrant and y is in third quadrant.

Since, in second quadrant, cos x and tan x are negative.

Since, in third quadrant, sec y is negative.

On using the formula,

cos x = – √(1 - sin² x)

tan x = sin x/cos x

Then,

cos x = – √(1 - sin² x)

= – √(1 – (3/5)²)

= – √(1 – 9/25)

= – √((25 - 9)/25)

= – √(16/25)

= – 4/5 

tan x = sin x/cos x

= (3/5)/(-4/5)

= 3/5 × -5/4

= -3/4

As we know that sec y = – √(1 + tan² y)

= – √(1 + (1/2)²)

= – √(1 + 1/4)

= – √((4 + 1)/4)

= – √(5/4)

= – √5/2

Then, 8 tan x – √5 sec y = 8(-3/4) – √5(-√5/2)

= -6 + 5/2

= (-12 + 5)/2

= -7/2

Thus, 8 tan x – √5 sec y = -7/2

C. sec θ = ½

Explanation:

According to the question,

We know that,

a) sin θ = – 1/5 is correct since Sin θ ∈ [-1,1]

b) cos θ = 1 is correct since Cos θ ∈ [-1,1]

c) sec θ = ½

⇒ (1/cos θ) = ½

⇒cos θ=2 is incorrect since Cos θ ∈ [-1,1]

d) tan θ = 20 is correct since tan θ ∈ R.

Thus, option (C) sec θ = ½ is the correct answer.

B. 1

Explanation:

According to the question,

tan 1° tan 2° tan 3°… tan 89°

= tan 1° tan 2° … tan 45° tan (90-44°) tan(90-43°)…tan (90-1°)

= tan 1°tan 2° … tan 45°cot 44°cot 43°…cot 1° [∵tan (90-θ)=cot θ]

= tan 1° cot 1° tan 2° cot 2°…tan45°… tan 89° cot 89°

=1.1….1 = 1

Thus, option (B) 1 is the correct answer.

Answer is (B) 

tan 1° tan 2° tan 3° … tan 89°

= [tan 1° tan 2° … tan 44°] tan 45°[tan (90° – 44°) tan (90° – 43°)… tan (90° – 1°)]

= [tan 1° tan 2° … tan 44°] [cot 44° cot 43°……. cot 1°]

=[tan 1°*cot 1°* tan 2°*cot 2°.....]   (tan 1°*cot 1° = 1)

= 1*1… 1*1 = 1

Answer is (D)

sin (45° + ) – cos (45° – ) = sin (45° + ) – sin (90° – (45° – ))

= sin (45° + ) – sin (45°+ ) = 0

LHS = (sin7x + sin x) + (sin5x + sin3x) 

= 2sin4xcos3x + 2sin4xcosx 

= 2sin4x(cos3x + cosx) 

= 2sin4x2cos2x cos x 

= 4cosxcos2xsin4x 

= RHS. 

LHS = cos2(2x) 

= 1- 2sin²2x = 1 - 2(2sinx cosx)² 

= 1- 8sin²xcos²x 

= RHS

LHS = sin \(\cfrac{13\pi}3\) sin \(\cfrac{2\pi}3\) + cos \(\cfrac{4\pi}3\) sin \(\cfrac{13\pi}6\)

= sin 780° sin 120° + cos 240° sin 390°

= sin (90° × 8 + 60°) sin (90° × 1 + 30°) + cos (90° × 2 + 60°) sin (90° × 4 + 30°)

We know that when n is odd, sin → cos.

= sin 60° cos 30° + [-cos 60°] sin 30°

= sin 60° cos 30° - sin 30° cos 60°

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

= sin (60° - 30°)

= sin 30°

= 1/2

= RHS

Hence proved.

LHS = sin \(\cfrac{13\pi}3\) sin \(\cfrac{8\pi}3\) + cos \(\cfrac{2\pi}3\) sin \(\cfrac{5\pi}6\) 

= sin 780° sin 480° + cos 120° sin 150°

= sin (90° × 8 + 60°) sin (90° × 5 + 30°) + cos (90° × 1 + 30°) sin (90° × 1 + 60°)

We know that when n is odd, cos → sin and sin → cos.

= sin 60° cos 30° + [-sin 30°] cos 60°

= sin 60° cos 30° - sin 30° cos 60°

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

= sin (60° - 30°)

= sin 30°

= 1/2

= RHS

Hence proved.