Explore topic-wise InterviewSolutions in .

L.H.S. = sec A (1 – sin A) (sec A + tan A)

= (sec A – sec A . sin A) (sec A + tan A) 

= (sec A – \(\frac{1}{cos\, A}\) . sin A) (sec A + tan A) 

= (sec A – tan A) (sec A + tan A) 

= sec² A – tan² A [∵ sec² A – tan² A = 1] 

= 1

Given ‘θ’ is acute => 0° < θ < 90° 

So sin 0° = 0 and sin 90° = 1 

So for 0° < θ < 90°, 

sin θ value lies in between zero and one. 

So sin θ value cannot be greater than 1. 

So sinθ = 5/3 does not exist.

L.H.S = \(\frac {cosec(90°-x)\, .sin(180°-x)\, .cot(360°-x)}{sec(180°+x)\, .tan(90°+x)\,.sin(-x)}\)

         = \(\frac{sec\,x\,sin\,x(-cot\,x)}{-sec\,x).(-cot\,x).(-sin\,x)}\)

= 1

= R.H.S

sin 90° =1 

cos 90° = 0 

tan 90° = not defined 

cot 90° = 0 

sec 90° = not defined 

cosec 90° = 1 

∴ tan 90°, sec 90° are not defined.

Given (sec² θ – 1) (cosec² θ – 1) 

= tan² θ × cot² θ [∵ sec² θ – tan² θ = 1; cosec² θ – cot² θ = 1] 

= tan² θ × \(\frac{1}{tan^2\,θ}\)

= 1

tanθ = \(\frac{24}7\)

⇒ sinθ = \(\frac{24}{25}\)

and cosθ = \(\frac{7}{25}\)

∴ sinθ + cosθ = \(\frac{24}{25}\) + \(\frac{7}{25}\) = \(\frac{31}{25}\)

Given sin A = \(\frac{3}{5}\), cos B = \(\frac{4}{5}\)

⇒ cos A = \(\frac{4}{5}\) & sin B = \(\frac{3}{5}\)

∵ sin²A + cos²A = 1 

(i) sin(A + B) = sinA cosB + cosA sinB 

= \(\frac{3}{5}\). \(\frac{4}{5}\) + \(\frac{4}{5}\).\(\frac{3}{5}\) = \(\frac{24}{25}\)

(ii) cos(A – B) = cosAcosB + sinA sinB

= \(\frac{4}{5}\). \(\frac{4}{5}\) + \(\frac{3}{5}\) .\(\frac{3}{5}\) = \(\frac{16}{25}\) + \(\frac{9}{25}\) = \(\frac{25}{25}\) = 1

Taking LHS = x² + y² + z²

Putting the values of x, y and z , we get

=(r cos α sin β)² + (r sin α sin β)² + (r cos α)²

= r² cos²α sin²β + r² sin²α sin²β + r² cos²α

Taking common r² sin² α , we get

= r² sin²α (cos²β + sin² β) + r²cos² α

= r² sin²α + r² cos² α [∵ cos² β + sin² β = 1]

=r² ( sin² α + cos² α)

= r² [∵ cos² α + sin² α = 1]

= RHS

Hence Proved

i. L.H.S = (sin² θ/cos θ) + cos θ 

= (sin² θ/cos² θ)/cos θ 

= 1/cos θ  ....[∵ sin² θ + cos² θ = 1]

= sec θ

= R.H.S

∵ (sin² θ/cos θ) + cos θ = sec θ

ii. L.H.S. = cos² θ (1 + tan² θ) 

= cos² θ sec² θ …[∵ 1 + tan² θ = sec² θ]

= cos² θ _^. 1/cos² θ  ....[∴ sec θ = 1/cos θ]

= 1

= R.H.S

∴ cos² θ (1 + tan² θ) = 1

i. sin 30° = 1/2

ii. cos 30° = √3/2

iii. tan 30° = 1/√3

iv. sin 60° = √3/2

v. cos 45° = 1/√2

vi. tan 45° = 1

(i) Sin 30°  = 1/2    (ii) cos30° = √3/2     (iii) tan30° = 1/√3   

(iv) sin60° = √3/2    (v) cos45° = 1/√2    (vi) tan45° = 1

i. sinθ/cosθ = [tanθ] 

ii. sinθ = cos (90 – θ)

iii. cosθ = (90 – θ) 

iv. tanθ × tan (90 – θ) = 1

cos 68° + tan 76° = cos (90° - 22°) + tan(90° - 22°) + tan (90° - 14°)

= sin 22° + cot 14°

[cos (90° - θ) = sin θ and tan  (90° - θ) = cot θ]

tan² 45° + cot² 30°

 = (1)² + (√3)² 

= 1 + 3 = 4

2 sin x = √3

sin x = \(\frac{\sqrt{3}}{2}\)

sin x = sin 60° 

∴ x = 60°

sin 3A = cos (A-26°) …(i)

We know that

Sin θ = cos (90° - θ)

So, Eq. (i) become

Cos (90° - 3A) = cos (A -26°)

On Equating both the sides, we get

90° - 3A = A – 26°

⇒ -3A - A = -26° -90°

⇒ -4A = -116°

⇒ A = 29°

cos 76° – sin 14°

We can write cos 76 as cos (90 – 14) 

∴ cos 76 = cos (90 – 14) = sin 14 

(∵ cos (90 – θ) = sin θ) 

∴ cos 76° – sin 14° 

= sin 14° – sin 14° 

= 0

(B) 0

According to the question,

We have to find the value of the equation,

cosec(75°+θ) – sec(15°-θ) – tan(55°+θ) + cot(35°-θ)

= cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]

Since, cosec (90°- θ) = sec θ

And, cot(90°-θ) = tan θ

We get,

= sec(15°-θ) – sec(15°-θ) – tan(55°+θ) + tan(55°+θ)

= 0

\(\frac{cos63°20'}{sin26°40'}\) = \(\frac{cos63°20'}{cos63°20'}\) = 1

(c) 2

  \(\frac{tan\,26°+tan\,19°}{x(1-tan\,26°tan\,19°)}\) = cos 60°

= \(\frac{tan\,26°+tan\,19°}{1-tan\,26°tan\,19°}\) = x cos 60°

= tan (26° + 19°) = \(x\) x \(\frac12\) \(\bigg[∵ tan (A + B)=\frac{tan\,A+tan\,B}{1-tan\,A\,tan\,B}\bigg]\)

= tan 45° = \(\frac{x}{2}\) ⇒ \(\frac{x}{2}\) = 1 ⇒ x = 2.

cos 19° 59′ + tan 12° 12′ + sin 49° 20′ = 0

cot(90° – 75) cot(90° – 60°) cot 45° cot 60° cot 75° 

= tan 75° tan 60° (1) cot 60° cot 75°

= 1

sin[π/3-sin^-1(-1/2)]

=sin[π/3-(-π/6)]

=sin π/2=1

In a right angled triangle ABC,

cos A = \(\frac{AC}{AB}\) and cos B = \(\frac{BC}{AB}\)

 ∵ cos A = cos B

\(\frac{AC}{AB}\) = \(\frac{BC}{AB}\)

∴ AC = BC

We have, opposite sides of equal angles are equal. 

Therefore, In a right angled triangle ABC

∠A = ∠B = 45°

Answer:

We have 
tan 2A = cot (A- 18°)
⇒ cot (90° - 2A) = cot (A -18°)
Equating angles,
⇒ 90° - 2A = A- 18° ⇒ 108° = 3A
⇒ A = 36°

Answer

In a triangle, sum of all the interior angles

A + B + C = 180°

⇒ B + C = 180° - A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

⇒ sin (B+C)/2 = sin (90°-A/2)

⇒ sin (B+C)/2 = cos A/2

⇒ sin(2θ + 45°) = cos(30°- θ)

⇒ sin(2θ + 45°) = sin{90°-(30° - θ)

⇒ 2θ + 45° = 90°- 30° + θ

⇒ θ = 15°

sinθ = cos(θ - 45°)

⇒ cos(90° - θ) = cos(θ - 45°)

⇒ 90° - θ = θ - 45°

⇒ 2θ = 135°

⇒ θ =\(\frac{135°}{2} = 67(\frac{1}{2})°\)

=2(\(\frac{sin^2\theta}{cos^2\theta}+\frac{1}{cos^2\theta}\))

= 2(\(\frac{sin^2\theta+1}{cos^2\theta}\))

=2 (\(\frac{1+ sin^2\theta}{1-sin^2\theta}\))

=R.H.S

= (sin² θ + cos² θ – sin θ cos θ) + (sin² θ + cos² θ + sinθ cosθ) 

= 2 (sin² θ + cos² θ) 

= 2(1) 

= 2 = R.H.S

LHS =  \(\cfrac{cos^3θ+sin^3θ}{cosθ+sinθ}\) + \(\cfrac{cos^3θ-sin^3θ}{cosθ-sinθ}\)

= \(\cfrac{(cosθ+sinθ)(cos^2θ-cosθsinθ+sin^2θ)}{cosθ+sinθ}\) + \(\cfrac{(cosθ-sinθ)(cos^2θ+cosθsinθ+sin^2θ)}{cosθ-sinθ}\)

= (cos² θ + sin² θ − cos θ sin θ) + (cos² θ + sin² θ + cos θ sin θ) 

= (1 − cos θ sin θ) + (1 + cos θ sin θ) 

= 2 

= RHS 

Hence, LHS = RHS

Correct option is (B) 20/21

cosec θ + cot θ = 5/2……(i) 

cosec² θ – cot² θ = 1

∴ (cosec θ – cot θ) (cosec θ – cot θ) = 1

 ∴ 5/2 (cosec θ – cot θ) = 1

∴ cosec θ – cot θ = 2/5...(ii)

Subtracting (ii) from (i), we get

2 cot θ = 5/2 - 2/5 = 21/10

∴ cot θ = 21/20

∴ tan θ = 20/21

Correct option is: B) 90°

Sin A = cos B

= sin A = sin (\(90^\circ\)-B) (\(\because\) cos \(\theta\) = sin (\(90^\circ\) - \(\theta\)) 

= A = \(90^\circ\) - B (By comparing angle)

= A + B = \(90^\circ\)

Correct option is: B) 90°

(i) LHS =  sin(70\(^\circ\) + θ) - cos(20\(^\circ\) - θ)

= sin\(\{90^\circ -(20^\circ-θ)\}\) - cos(20\(^\circ\) - θ)

= cos(20\(^\circ\)- θ) - cos(20\(^\circ\) - θ)

= 0

= RHS

(ii) LHS = tan(55\(^\circ\)- θ} - cot(35\(^\circ\)+ θ

= tan\(\{90^\circ -(35^\circ-θ)\}\) - cos(35\(^\circ\) - θ)

= cot (35\(^\circ\)- θ) - cot(35\(^\circ\) - θ)

= 0

= RHS

(iii) LHS =  cosec(67\(^\circ\)+ θ) - sec(20\(^\circ\)- θ)

= cosec\(\{90^\circ -(23^\circ-θ)\}\) - sec(23\(^\circ\) - θ)

= sec (23\(^\circ\)- θ) - sec(23\(^\circ\) - θ)

= 0

= RHS

(iv) LHS =  cosec(65\(^\circ\)+ θ) - sec(25\(^\circ\)- θ) - tan(55\(^\circ\)- θ) + cot(35\(^\circ\)+ θ)

= cosec\(\{90^\circ -(25^\circ-θ)\}\) - sec(25\(^\circ\) - θ) - tan(55\(^\circ\)- θ) + cot\(\{90^\circ -(55^\circ-θ)\}\)

= sec (25\(^\circ\)- θ) - sec(25\(^\circ\) - θ) - tan(55\(^\circ\) - θ) + tan(55\(^\circ\) - θ)

= 0

= RHS

(v) LHS = sin(50\(^\circ\)+ θ) - cos(40\(^\circ\)- θ) + tan1\(^\circ\)tan 10\(^\circ\)tan80\(^\circ\) tan89\(^\circ\)

= sin \(\{90^\circ -(40^\circ-θ)\}\) - cos (40\(^\circ\)- θ) + \(\{tan1^\circ tan(90^\circ-1^\circ)\}\)\(\{tan10^\circ tan(90^\circ-10^\circ)\}\)

= cos (40\(^\circ\) - θ) - cos(40\(^\circ\) - θ) + (tan1\(^\circ\)cot1\(^\circ\))(tan10\(^\circ\)cot10\(^\circ\))

= \((\cfrac{1}{cot1^\circ}\times{cot1^\circ})\)\((tan10^\circ\times\cfrac{1}{tan10^\circ})\)

= 1 x 1

= 1

= RHS

(i) LHS =  tan 5\(^\circ\) tan 25\(^\circ\)tan30\(^\circ\)tan 65\(^\circ\) tan 85\(^\circ\)= 1

tan (90\(^\circ\)- 85\(^\circ\)) tan( 90\(^\circ\)- 65\(^\circ\)) x \(\frac{1}{\sqrt3}\) x \(\cfrac{1}{cot60^\circ}\) \(\cfrac{1}{cot85^\circ}\)

cot 85\(^\circ\)cot 65\(^\circ\)\(\frac{1}{\sqrt3}\)  \(\cfrac{1}{cot60^\circ}\) \(\cfrac{1}{cot85^\circ}\)

= \(\frac{1}{\sqrt3}\) = RHS

(ii) LHS =  cot12\(^\circ\) cot 38\(^\circ\)cot 52\(^\circ\) cot 60\(^\circ\)cot78\(^\circ\)

= tan(90\(^\circ\) - 12\(^\circ\)) x tan (90\(^\circ\) - 38\(^\circ\)) x cot52\(^\circ\)x \(\frac{1}{\sqrt3}\)x cot 78\(^\circ\)

= \(\frac{1}{\sqrt3}\) x tan 78\(^\circ\)x tan 52\(^\circ\)x cot52\(^\circ\)x   \(\cfrac{1}{tan52^\circ}\) \(\cfrac{1}{tan78^\circ}\)

= \(\frac{1}{\sqrt3}\)

= RHS

(iii) LHS =  cos 15\(^\circ\) cos 35\(^\circ\) cosec55\(^\circ\) cos 60\(^\circ\)cosec 75\(^\circ\) 

cos(90\(^\circ\)- 75\(^\circ\))cos (90\(^\circ\)- 55\(^\circ\))\(\cfrac{1}{sin55^\circ}\)x \(\frac{1}2\)x\(\cfrac{1}{sin75^\circ}\)

= sin75\(^\circ\)sin55\(^\circ\)\(\cfrac{1}{sin55^\circ}\) x \(\frac{1}2\)x\(\cfrac{1}{sin75^\circ}\)

= \(\frac{1}{2}\)

= RHS

(iv) LHS = cos1\(^\circ\) cos 2\(^\circ\) cos3\(^\circ\)......cos 180\(^\circ\)

= cos1\(^\circ\)x cos 2\(^\circ\)x cos3\(^\circ\)x......x cos 90\(^\circ\)x....x cos180\(^\circ\)

= cos1\(^\circ\)x cos 2\(^\circ\)x cos3\(^\circ\)x......x 0 x....x cos180\(^\circ\)

= 0

= RHS

(v) \((\cfrac{sin49^\circ}{cos41^\circ})\) + \((\cfrac{cos41^\circ}{sin49^\circ})\)= 2

= \((\cfrac{cos(90^\circ-49^\circ)}{cos41^\circ})^2\) + \((\cfrac{cos41^\circ}{​​cos(90^\circ-49^\circ)})^2\)

=   \((\cfrac{cos41^\circ}{cos41^\circ})^2\) + \((\cfrac{cos41^\circ}{cos41^\circ})^2\)

= 1² + 1² 

= 1 + 1

= 2

= RHS 

2/3cosec²58° - 2/3cot58°tan32° - 5/3 tan13°tan37°tan 45°tan 53°tan 77°

= 2/3(cosec²58°- cot 58°tan32°) - 5/3 tan13°tan(90° - 13°) tan 37°tan(90° - 37°)(tan45°)

= 2/3[cosec²58°- cot 58°tan(90°- 58°)] - 5/3 tan13°cot13°tan37°cot37°(1)

= 2/3(cosec²58°- cot58°tan58°)- 5/3tan13°\(\frac{1}{tan13°}\)tan37°\(\frac{1}{tan37°}\)

= 2/3(cosec²58°- cot²58°- 5/3

= 2/3 - 5/3

= - 1

Hence proved

As, sin θ = 1/2 

so, cosecθ = 1/sin θ = 2 … . . (i)

Now, 

3 cot² θ + 3 

= 3(cot² θ + 1) 

= 3cosec²θ 

= 3(2)²      [Using (i)] 

= 3(4) 

= 12

3cot²θ - 3cosec²θ

= 3(cot² θ − cosec²θ) 

= 3(−1) 

= -3

We have, 

\(\frac{tan 10°}{cot 80°}\)

= \(\frac{tan (90° – \,10°) }{cot 80°}\)

= \(\frac{cot 80°}{cos 80° }\)

= 1 [∵ tan (90 – θ) = cot θ]

Given sin 5° cos 85° + cos 5° sin 85° 

= sin 5° . cos (90° – 5°) + cos 5° . sin (90° – 5°) 

= sin 5° . sin 5° + cos 5° . cos 5° 

[∵ sin (90° – θ) = cos θ; cos (90° – θ) = sin θ] 

= sin² 5° + cos² 5° 

= 1 [∵ sin² θ + cos² θ = 1]

tan (A – B)= 1/√3

or tan(A – B) = tan 30°

A – B = 30° …(1)

Again, tan(A+B) = √3

= tan 60°

A + B = 60° …(2)

Solving (1) and (2), we get

2A = 90° or A = 45°

Putting A = 45° in (1), we get

45° – B = 30° or B = 45° – 30° = 15°

Therefore, A = 45°, B = 15°

Correct answer = (d) 1

\(\cfrac{sec30^\circ}{cosec60^\circ} \) = \(\cfrac{sec30^\circ}{sec(90^\circ-60^\circ)} \) = \(\cfrac{sec30^\circ}{sec30^\circ} \) = 1

Given: sec5 A = cosec(A - 36°)

=> cosec(90° − 5A) = cosecA − 36° ) [∵ cosec(90° − θ) = sec θ] 

=> 90° − 5A = A − 36° 

=> 6A = 90° + 36° 

=> 6A = 126° 

=> A = 21°