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Concept:

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

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

Calculation:

Here, we have to find the value of sin 75° sin 15° + cos 75° cos 15°

As we know that, cos x cos y + sin x sin y = cos (x - y)

∴ sin 75° sin 15° + cos 75° cos 15° = cos 75° cos 15° + sin 75° sin 15°

= cos (75° - 15°)

= cos 60° 

= \(\frac {1}{2}\)

Concept:

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

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

Calculation:

Here, we have to find the value of sin 75° cos 15° + cos 75° sin 15°

As we know that, sin x cos y + cos x sin y = sin (x + y)

∴ sin 75° cos 15° + cos 75° sin 15° = sin (75° + 15°)

= sin 90° 

= 1

Concept:

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

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

Calculation:

Here, we have to find the value of sin 75° cos 15° - cos 75° sin 15°

As we know that, sin x cos y - cos x sin y = sin (x - y)

∴ sin 75° cos 15° - cos 75° sin 15° = sin (75° - 15°)

= sin 60° 

= \(\frac {\sqrt 3}{2}\)

Concept:

Inverse trigonometric identity

• \(\rm \tan^{-1}{a} + \tan^{-1}b = \tan^{-1}{\left(a+b\over1-ab\right)}\)
• \(\rm \tan^{-1}{a} - \tan^{-1}b = \tan^{-1}{\left(a-b\over1+ab\right)}\)
• sin^-1(sin a) = a
• cos-1(cos a) = a
• tan-1(tan a) = a
• cot-1(cot a) = a

Calculation:

S = \(\rm \cot\left[\tan^{-1}{a} - \tan^{-1}{\left({x-y\over x+y}\right)}\right] = 1\)

⇒ \(\rm \cot\left[\tan^{-1}{a} - \tan^{-1}{\left({x-y\over x+y}\right)}\right] = \cot45\) (∵ cot 45 =1)

Taking cot-1 on both sides

⇒ \(\rm \cot^{-1}\cot\left[\tan^{-1}{a} - \tan^{-1}{\left({x-y\over x+y}\right)}\right] = \cot^{-1}(\cot45)\)

⇒ \(\rm \tan^{-1}{a} - \tan^{-1}{\left({x-y\over x+y}\right)} = 45\)

⇒ \(\rm \tan^{-1}\left[{a - {\left({x-y\over x+y}\right)}\over1+a\times{\left({x-y\over x+y}\right)}}\right] = 45\)

Taking tan on both sides

⇒ \(\rm \tan \tan^{-1}\left[{a - {\left({x-y\over x+y}\right)}\over1+a{\left({x-y\over x+y}\right)}}\right] = \tan45\)

⇒ \(\rm {a - {\left({x-y\over x+y}\right)}\over1+a{\left({x-y\over x+y}\right)}} = 1\)

⇒ \(\rm {a - {\left({x-y\over x+y}\right)}=1+a{\left({x-y\over x+y}\right)}}\)

⇒ \(\rm {a {\left(1-{x-y\over x+y}\right)}=1+{\left({x-y\over x+y}\right)}}\)

⇒ a(x + y - (x - y)) = x + y + x - y

⇒ 2ya = 2x

⇒ a = \(\boldsymbol{\rm x\over y}\)

Concept:

Inverse trigonometric identities

• sin-1(sin a) = a
• cos-1(cos a) = a
• tan-1(tan a) = a
• cot-1(cot a) = a
• cot^-1 a = 90 - tan^-1 a
• cos-1 a = 90 - sin-1 a

Trigonometric Identities

• cos θ = sin (90 - θ)
• cot θ = tan (90 - θ)

Calculation:

\(\rm \sin\left(\tan^{-1}X\right) = \cos\left(\cot^{-1}{1\over5}\right)\)

⇒ \(\rm \sin\left(\tan^{-1}X\right) = \sin\left[90-\left(\cot^{-1}{1\over5}\right)\right]\)

⇒ \(\rm \tan^{-1}X = 90-\cot^{-1}{1\over5}\)

⇒ \(\rm \tan^{-1}X = 90-\left(90-\tan^{-1}{1\over5}\right)\)

⇒ \(\rm \tan^{-1}X = \tan^{-1}{1\over5}\)

Taking tan both sides

⇒ \(\rm \tan (\tan^{-1}X) = \tan(\tan^{-1}{1\over5})\)

⇒ X = \(\boldsymbol{\rm 1\over5}\) 

√((1 – A²)/(B² – A²))

Given

Cot x/Cot y = A and Cosec x/Cosec y = B

Concept

Cosec² A – Cot²A = 1

Calculation

Cot x = A Cot y

Squaring on both sides

Cot²x = A²Cot²y

⇒  Cosec ²x – 1 = A ² (Cosec ² y – 1)     ---- (1)

Cosec x = B Cosec y

Squaring on both sides

⇒  Cosec ²x = B ²Cosec ² y    ---- (2)

From (1) and (2)

B²Cosec²y – 1 = A²Cosec²y – A²

⇒  B²Cosec²y – A²Cosec²y = - A² + 1

⇒  Cosec²y(B²– A²) = - A²+ 1

Given

Sec A + Tan A = 4

Concept

Sec² A – Tan²A = 1

(a - b)² = a² - 2ab + b²

Calculation

Sec = 4 – Tan A

Squaring on both sides

Sec²A = 16 + Tan²A – 8 Tan A

⇒  1 + Tan²A = 16 + Tan²A – 8 Tan A

⇒  1 – 16 = - 8 Tan A

⇒  15/8 = Tan A

So, Cot A = 8/15

Now, [(8/15) + 1]/[(8/15) – 1]

⇒   [(8 + 15)/15]/[(8 – 15)/15]

Given

A = 90 °

Concept

Cos 90 ° = 0

Calculation

3/2(√1 + Cos 90°) + ½(√1 – Cos 90°)

⇒  3/2(√1 + 0) + ½(√1 – 0)

⇒  3/2 + ½

⇒  2 = Sec 60°

Given:

tan(x + y) × tan(x – y) = 1

Concept used:

If tanA × tanB = 1, then A + B = 90°

Calculation:

tan(x + y) × tan(x – y) = 1

⇒ x + y + x – y = 90°

⇒ 2x = 90°

⇒ x = 45°

⇒ tan45° = 1

Concept:

• sin2 θ + cos2 θ = 1.
• (a ± b)2 = a2 ± 2ab + b2.

Calculation:

sin x + a cos x = b

⇒ (sin x + a cos x)2 = b2

⇒ sin2 x + a2 cos2 x + 2 a sin x cos x = b2

⇒ (1 - cos2 x) + a2 cos2 x + 2a sin x cos x = b2            ... [Using sin2 x = 1 - cos2 x]

⇒ (a2 - 1) cos2 x + 2a sin x cos x = b2 - 1

⇒ 2a sin x cos x = b2 - 1 + (1 - a2) cos2 x            ... (1)

Let k = |a sin x - cos x|

⇒ k2 = (a sin x - cos x)2

⇒ k2 = a2 sin2 x + cos2 x - 2a sin x cos x

⇒ k2 = a2 (1 - cos2 x) + cos2 x - 2a sin x cos x            ... [Using sin2 x = 1 - cos2 x]

⇒ k2 = a2 + (1 - a2) cos2 x - 2a sin x cos x

⇒ k2 = a2 + (1 - a2) cos2 x - [b2 - 1 + (1 - a2) cos2 x]            ... [Using equation (1)]

⇒ k2 = a2 - b2 + 1

⇒ \(\rm k = \sqrt{a^2 - b^2 + 1}\).

∴ \(\rm |a \sin x-\cos x| = \sqrt{a^2 - b^2 + 1}\).

Given:

\(\sqrt {2\; + \;\sqrt {2\; + \;\sqrt {2\; + \;2{\rm{cos}}16{\rm{\theta }}} \;} \;} \)

Concept used:

\(\sqrt {2\; + \;\sqrt {2\; + \;\sqrt {2\; + \;2{\rm{cos}}16{\rm{\theta }}} \;} \;} = \;2{\rm{cos}}\frac{{16\theta }}{{{2^{\rm{n}}}}}\)

Where n → number of terms

Calculation:

Here 2 came 3 times

⇒ n = 3

\(\sqrt {2\; + \;\sqrt {2\; + \;\sqrt {2\; + \;2{\rm{cos}}16{\rm{\theta }}} \;} \;} = \;2{\rm{cos}}\frac{{16\theta }}{{{2^3}}}\)

⇒ 2cos2θ

\(\therefore {\bf{The}}\;{\bf{value}}\;{\bf{of}}\;\sqrt {2\; + \;\sqrt {2\; + \;\sqrt {2\; + \;2{\bf{cos}}16{\bf{\theta }}} \;} \;} {\bf{is}}\;2{\bf{cos}}2{\bf{\theta }}.\)

Given:

sinθ = 3/5

Concept used:

In a right-angled triangle

sinθ = Perpendicular/hypotenuse

tanθ = perpendicular/base

Pythagoras theorem

H² = P² + B²

Calculation:

H² = P² + B²

⇒ B² = 5² - 3²

⇒ B² = 25 – 9

⇒ B² = 16

⇒ B = 4

⇒ tanθ = 3/4

∴ The value of tanθ is 3/4.

Given

sin x – cos x = 0

Calculation

sin x – cos x = 0

⇒ sinx = cosx

⇒ sinx = sin(90° - x)

⇒ x = 90° - x

⇒ 2x = 90°

⇒ x = 45°

Now,

(sec x + cosec x)2 

= (sec 45° + cosec 45°)2

= (√2 + √2)²

= (2√2)2

= 8

Given

Sin x = 0

Concept

Sin 0°= 0

Sec 0° = 1

Cos 0 ° = 1

Calculation

Sin x = Sin 0°

x = 0

So, Sec(0) + Cos(0)

⇒  1 + 1

Given:

√3 - 3√3tan2A = 3tan A - tan3A

Formula used:

tan 3A = (3tan A - tan3A)/(1 - 3tan²A)

Calculation:

√3 - 3√3tan2A = 3tan A - tan3A

⇒ √3(1 - 3tan2A) = 3tan A - tan3A

⇒ √3 = (3tan A - tan3A)/(1 - 3tan2A)

⇒ √3 = tan 3A 

⇒ tan 60° = tan 3A

⇒ 3A = 60° 

⇒ A = 60°/3 = 20° 

∴ The value of A is 20°. 

Given:

Sin(A + B) = Sin A + Sin B

Formula Used:

Sin(A + B) = SinA.CosB + CosA.SinB

Calculation:

From option 1,

A = 0° and B = 90° 

We have,

⇒ Sin(0° + 90° ) = Sin 0° + Sin 90° 

⇒ Sin (90°) = 0 + 1

⇒ 1 = 1

∴ The value of A and B are 0° and 90° respectively.

Calculation:

Given: a sin A + b cos A = c

Divide both sides by \(\rm \frac {1}{\sqrt {a^2 +b^2}}\), we get

⇒ \(\rm \frac {a}{\sqrt {a^2 +b^2}}\) sin A + \(\rm \frac {b}{\sqrt {a^2 +b^2}}\) cos A = \(\rm \frac {c}{\sqrt {a^2 +b^2}}\)

Let sin α = \(\rm \frac {a}{\sqrt {a^2 +b^2}}\) and cos α = \(\rm \frac {b}{\sqrt {a^2 +b^2}}\)

⇒ sin A sin α + cos A cos α = \(\rm \frac {c}{\sqrt {a^2 +b^2}}\)

⇒ cos (A - α) = \(\rm \frac {c}{\sqrt {a^2 +b^2}}\)

⇒ A - α = cos^-1 \(\rm \frac {c}{\sqrt {a^2 +b^2}}\)

⇒ A = scos-1 \(\rm \frac {c}{\sqrt {a^2 +b^2}}\) + α

Now, \(\rm \tan α = \frac {\sin α}{\cos α} = \frac ab\)

∴ α = tan^-1 \(\rm \frac ab\)

So, A = cos-1 \(\rm \frac {c}{\sqrt {a^2 +b^2}}\) + tan-1 \(\rm \frac ab\) = \(\rm \tan^{-1} \left(\frac{a}{b}\right) + \cos^{-1} \left(\frac{c}{\sqrt{a^2+b^2}}\right)\)

Concept:

Trigonometric Identities:

sin² θ + cos² θ = 1.

sin 2θ = 2 sin θ cos θ.

cos 2θ = cos² θ - sin² θ.

Calculation:

Let us express \(\rm\dfrac{\sin x}{1 + \cos x}\) in terms of tan x.

\(\rm \dfrac{\sin x}{1 + \cos x}=\dfrac{2\sin \tfrac{x}{2}\cos \tfrac{x}{2}}{\left (\cos^2 \tfrac{x}{2}+\sin^2 \tfrac{x}{2} \right ) + \left (\cos^2 \tfrac{x}{2}-\sin^2 \tfrac{x}{2} \right )}\)

= \(\rm \dfrac{2\sin \tfrac{x}{2}\cos \tfrac{x}{2}}{2\cos^2 \tfrac{x}{2}}= \dfrac{\sin \tfrac{x}{2}}{\cos \tfrac{x}{2}}=\tan \dfrac{x}{2}\).

∴ \(\rm f(x)=\tan^{-1} \left[\dfrac{\sin x}{1 + \cos x}\right]= \tan^{-1}\left (\tan \dfrac{x}{2} \right )=\dfrac{x}{2}\).

And, the first derivative of f(x) = f'(x) = \(\rm \dfrac{d}{dx}\left (\dfrac{x}{2} \right )=\dfrac{1}{2}\).

Concept:

Double angle formula:

\(\rm \sin2x=\dfrac{2\tan x}{1+\tan^2}\)

Addition formula:

\(\rm \tan(x+y) = \dfrac{\tan x + \tan y}{1-\tan x\tan y}\)

Calculation:

The given identity is \(\rm \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \sin^{-1}\left(\frac{2b}{1+b^2}\right) = 2\tan^{-1}n\).

Let \(\rm a = \tan y_1\mbox{ and } b= \tan y_2\).

Therefore, the given equation becomes:

\(\rm \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \sin^{-1}\left(\frac{2b}{1+b^2}\right)= 2\tan^{-1}n\)

\(\rm \sin^{-1}\left(\frac{2(\tan y_1)}{1+(\tan y_1)^2}\right) + \sin^{-1}\left(\frac{2(\tan y_2)}{1+(\tan y_2)^2}\right) = 2\tan^{-1}n\)

\(\rm \sin^{-1}\left(\sin 2y_1\right)+\sin^{-1}\left(\sin 2y_2\right) = 2\tan^{-1}n\)

\(\rm 2y_1 +2y_2 = 2\tan^{-1}n\)

\( \rm y_1 +y_2 = \tan^{-1}n \)

\(\rm \tan(y_1 + y_2) = n\)

\(\rm \frac{\tan y_1 + \tan y_2}{1-\tan y_1\tan y_2}=n\)

\(\rm \frac{a+b}{1-ab}=n \)

Therefore, \(\rm n = \frac{a+b}{1-ab}\).

Concept

Sin ²A + Cos ²A = 1

(a + b)² = a²+ b²+ 2ab

(a³ + b³)= (a + b) (a² – ab + b²)

Calculation

(Sin²A)³ + (Cos² A)³/(Sin²A + Cos²A) (1 - Sin A Cos A)

⇒ (sin²A) + (Cos²A) [(Sin⁴A) – Sin² A Cos² A + (Cos ⁴A)]/(Sin ²A + Cos ²A) (1 - Sin A Cos A)

⇒ (sin² A + Cos²A) [(Sin⁴ A + Cos⁴ A – Sin² A Cos² A)]/(Sin ²A + Cos ² A) (1 - Sin A Cos A)

⇒ (Sin ²A + Cos ² A) [(1 - Sin A Cos A)]/(Sin ²A + Cos ²A) (1 -  Sin A Cos A)

Given:

cosecθ = 3x 

cotθ = 3/x

Formula used:

cosec²θ - cot²θ = 1

Calculation:

cosecθ = 3x 

⇒ cosec²θ = 9x²               [Squaring both sides]       .......(1)

cotθ = 3/x

⇒ cot²θ = 9/x²                  [Squaring both sides]      ..........(2)

Subtract (2) in (1)

cosec2θ - cot2θ = 9x2 - 9/x2 

⇒ 1 = 9(x2 - 1/x²)

⇒ (x2 - 1/x2) = 1/9

∴ The value of x2 - 1/x² is 1/9.

Given:

cotθ = 4/3

Formula used:

cosec²θ  – cot²θ = 1

Calculation:

cotθ = 4/3

⇒ cosec²θ  – cot²θ = 1

⇒ cosec²θ = 1 + cot²θ

⇒ cosec²θ = 1 + (4/3)²

⇒ cosec²θ = 1 + (16/9)

⇒ cosec²θ = (25/9)

⇒ cosecθ = √(25/9)

⇒ cosecθ = 5/3

Given:

Value of sin(A – B) = 0

Identity used:

sin²A + cos²A = 1

Calculation:

As, sin(A – B ) = 0

⇒ sin(A – B ) = sin0° 

⇒ A – B = 0

⇒ A = B

We have to find the value of 2sinA × sinB + 2cosA × cosB 

⇒ 2sinA × sinB + 2cosA × cosB  = 2 × (sinA × sinA + cosA × cosA)

⇒  2 × (sin²A + cos²A)

⇒ 2

Now 2 is also the even prime number less than 5.

∴ Both options 1 and 2 are correct.

Given:

cos 2θ = 0.28

Concepts used:

cos 2θ = 2 cos²θ - 1

Pythagoras theorem:

H2 = P2 + B² 

cosθ = B/H, cosecθ = H/P, tanθ = P/B, sinθ = P/H

Calculation:

⇒ cos 2θ = 2 cos2θ - 1

⇒ 0.28 = 2 cos2θ - 1

⇒ 2 cos2θ = 0.28 + 1

⇒ 2 cos2θ = 1.28

⇒ cos2θ = 0.64

⇒ cosθ = √0.64

⇒ cosθ = 0.8/1

As cosθ = Base/hypotenuse

⇒ 0.8 = Base/hypotenuse

On comparing, Base (B) = 0.8, hypotenuse (H) = 1 

Using Pythagoras theorem,

H2 = P2 + B² 

⇒ 1² = P² + (0.8)² 

⇒ 1 = P² + 0.64

⇒ 1 - 0.64 = P²

⇒ P² = 0.36

⇒ P² = √0.36

⇒ P = 0.6

Value of sinθ = P/H

⇒ 0.6/1

⇒ 0.6

Value of cosecθ = H/P

⇒ 1/0.6

⇒ 1.67

Value of tanθ = P/B

⇒ 0.6/0.8

⇒ 0.75

Value of expression, (cosecθ - tanθ + sinθ)/(cosecθ + tanθ + sinθ) = (1.67 - 0.75 + 0.6)/(1.67 + 0.75 + 0.6)  

⇒ 1.52/3.02

⇒ 0.503

∴ The value of the expression is 0.503.

Concept:

Trigonometric Identities:

• \(\rm \tan\theta=\dfrac{\sin\theta}{\cos\theta}\).
• \(\rm \cot\theta=\dfrac{\cos\theta}{\sin\theta}\).
• sin² θ + cos² θ = 1.
• sin 2θ = 2 sin θ cos θ.
• cos 2θ = cos² θ - sin² θ.

Calculation:

Let us observe that:

\(\rm \cot 2\theta=\dfrac{\cos 2\theta}{\sin2\theta}=\dfrac{\cos^2\theta-\sin^2\theta}{2\sin\theta\cos\theta}=\dfrac{1}{2}(\cot\theta-\tan\theta)\)

⇒ cot θ - tan θ = 2 cot 2θ             ... (1)

⇒ tan θ = cot θ - 2 cot 2θ             ... (2)

Now, tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ

= (cot θ - 2 cot 2θ) + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ             ... Using equation (2)

= cot θ - 2(cot 2θ - tan 2θ) + 4 tan 4θ + 8 cot 8θ

= cot θ - 2(2 cot 4θ) + 4 tan 4θ + 8 cot 8θ             ... Using equation (1)

= cot θ - 4(cot 4θ - tan 4θ) + 8 cot 8θ

= cot θ - 4(2 cot 8θ) + 8 cot 8θ             ... Using equation (1)

= cot θ - 8 cot 8θ + 8 cot 8θ

= cot θ.

• sin (A ± B) = sin A cos B ± sin B cos A.
• cos (A ± B) = cos A cos B ∓ sin A sin B.

GIVEN:

2sin10°.sin50°.sin70°

FORMULA:

(sin3θ)/4 = sin θ. sin(60° - θ).sin(60° + θ)

CALCULATION:

2sin10°.sin50°.sin70°

⇒ 2sin10°.sin(60° - 10°).sin(60° + 10°)

⇒ 2(sin3(10°))/4

⇒ (sin30°)/2

⇒ (1/2)/2

⇒ 1/4

∴ The value is 1/4.

GIVEN:

X = (1 + tan42°)(1 + tan3°)

FORMULA:

If A = B + C

tan A = tan(B + C) = (tan B + tan C)/(1 – tan B tan C)

⇒ tan A = (tan B + tan C)/(1 – tan B tan C)

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

⇒ tan A = tan B + tan C + tan A tan B tan C

CALCULATION:

X = (1 + tan42°)(1 + tan3°)

⇒ X = 1 + tan42° + tan3° + tan42°tan3°

⇒ X = 1 + (tan42° + tan3° + tan42°tan3°)

⇒ X = 1 + (tan42° + tan3° + tan45°tan42°tan3°)

⇒ X = 1 + tan45°

⇒ X = 1 + 1

⇒ X = 2

GIVEN:

\(\frac{{tan35^\circ \; + \;tan25^\circ \; + \;√ 3 tan25^\circ tan35^\circ }}{{\left( {tan19^\circ \; + \;tan26^\circ \; + \;tan19^\circ tan26^\circ } \right)}}\)

FORMULA:

If A = B + C, then

tan A = tan(B + C) = (tan B + tan C)/(1 – tan B tan C)

⇒ tan A = (tan B + tan C)/(1 – tan B tan C)

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

⇒ tan A = tan B + tan C + tan A tan B tan C

CALCULATION:

For Numerator:

tan35° + tan25° + √3tan25°tan35° = tan35° + tan25° + tan60°tan25°tan35°

⇒ tan60°

⇒ √3

For Denominator:

tan19° + tan26° + tan19°tan26° = tan19° + tan26° + tan45°tan19°tan26°

⇒ tan45°

⇒ 1

 \(\frac{{tan35^\circ \; + \;tan25^\circ \; + \;√ 3 tan25^\circ tan35^\circ }}{{\left( {tan19^\circ \; + \;tan26^\circ \; + \;tan19^\circ tan26^\circ } \right)}}\) = √3/1 

⇒ √3

∴ The value is √3.

Calculation:

sin(θ + 30°) = 3/√12

⇒ sin(θ + 30°) = 3/2√3      ----(√12 = 2√3)

⇒ sin(θ + 30°) = (3/2√3) × (√3/√3)

⇒ sin(θ + 30°) = (3/2√3) × (√3/√3)

⇒ sin(θ + 30°) = √3/2

⇒ sin(θ + 30°) = sin 60° 

⇒ (θ + 30°) = 60° 

⇒ θ = 30° 

The value of θ is equal to 30°

Given:

\(\frac{{secθ + tanθ }}{{secθ - tanθ }} = 5\)

Concept used:

secθ = Hypotenuse/Base, tanθ = perpendicular/base, cosθ = base/hypotenuse

Pythgorus theorem

Hypotenuse² = Perpendicular² + Base²

Calculation:

Let Perpendicular = P, Base = B, Hypotenuse = H

\(\frac{{secθ + tanθ }}{{secθ - tanθ }} = 5\)

⇒ {(H/B) + (P/B)}/{(H/B) - (P/B)} = 5/1

⇒ (H + P)/(H - P) = 5/1

⇒ H + P = 5      ----(1)

⇒ H - P = 1      ----(2)

Solve (1) and (2)

⇒ H = 3, P = 2

Hypotenuse2 = Perpendicular2 + Base2

⇒ Base = √5

\(\frac{{3{{\cos }^2}\theta + 1}}{{3{{\cos }^2}\theta - 1}}\)

⇒ {3 × (√5/3)² + 1}/{3 × (√5/3)2 - 1}

⇒ (8/3)/(2/3)

⇒ 4

∴ The value is 4.

Given:

sin(A – B) = 1/2

Calculation:

We know that

sin 30° = 1/2

sin(A – B) = sin 30° 

⇒ A – B = 30°      ----(i)

Also, cos(A + B) = 1/2

cos 60° = 1/2

cos(A + B) = cos 60° 

⇒ A + B = 60°      ----(ii)

Adding (i) and (ii) 

2A = 90° 

⇒ A = 45° 

∴ The value of A is 45° 

Given:

sinx = 3/5

Formula Used:

sin²θ + cos²θ = 1

Calculations:

3sin2x + 5cos2x + 2

⇒ 3sin2x + 5(1 - sin2x) + 2 = 7 - 2sin2x 

Now, sinx = 3/5

⇒ 7 - 2sin2x = 7 - 2 × (3/5)²

⇒ 7 - 2sin2x = 157/25

∴ The value of (3sin2x + 5cos2x + 2) is 157/25.

Trigonometry identity used:

Sin²θ + cos²θ = 1

Calculation:

12 sin²θ + 13 cos²θ

= 12 sin²θ + 12 cos²θ + cos²θ

= 12 (sin²θ + cos²θ) + cos²θ

= 12 + cos²θ

For minimum value,

Minimum value of cosθ = –1

But cos²θ  ≥ 0, when θ  = 90°

So, cos0° = 1,

Then, required minimum value

= 12 + 0

= 12.

For the maximum value,

Maximum value of cosθ = 1

And cos²θ =1

Then, required maximum value,

= 12 + 1 = 13

∴The minimum and maximum values of 12 sin²θ +13 cos²θ are 12 and 13.

Given :-

ΔABC is a right angle at B

sin A = (1/√2)

Concept :-

As sin A = (1/√2)

sin A = sin45° 

A = 45° 

Calculation :-

As B is right angle and,

⇒ ∠A = 45° 

Sum of triangle = 180° 

⇒ ∠A + ∠B + ∠C = 180° 

⇒ 45° + 90° + ∠C = 180° 

⇒ ∠C = 180° - 135° 

⇒ ∠C = 45° 

As ∠A = ∠C = 45° 

⇒ sin A = cos C = cos A = (1/√2)

ΔABC is an isosceles triangle 

Now, Put the value of all identities 

⇒ \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) = (sin A (sin A + sin A))/(sin A (sin A + sin A))

⇒ \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) = 1

∴  \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) = 1

Given :

\(\frac{sec A(sec A + tan A )(1 - sin A)}{(cosec^2 A - 1) sin^2 A}\)

Concept used :

sec a = 1/(cos a)

tan a = (sin a)/(cos a)

Solution :

\(\frac{sec A(sec A + tan A )(1 - sin A)}{(cosec^2 A - 1) sin^2 A}\)

\( = \left[ {\frac{1}{{cosA}}\left( {\frac{1}{{cosA}} + \frac{{sinA}}{{cosA}}} \right)\left( {1 - sinA} \right)} \right]/[(1/{\sin ^2}A) - 1]\;{\rm{sin}}{\;^2}A\;\)

\( = [1/cosA(1 + sinA)(1 - sinA)]/[((1 - si{n^2}A)/si{n^2}A){\rm{ \times }}si{n^2}A]\)

\( = se{c^2}A(1 - si{n^2}A)/((1 - si{n^2}A))\)

= sec² A

Given :

\(\frac{(1 + tanθ + secθ)(1 + cotθ - cosecθ)}{(secθ + tanθ)(1 - sinθ)}\)

Concept used :

tanθ = sinθ/cosθ

secθ = 1/cosθ 

cotθ = cosθ\sinθ

Solution :

 \(\frac{(1 + tanθ + secθ)(1 + cotθ - cosecθ)}{(secθ + tanθ)(1 - sinθ)}\)

\( = \;\frac{{\left( {1 + \frac{{sinθ }}{{cosθ }} + \frac{1}{{cosθ }}} \right)\left( {1 + \frac{{cosθ }}{{sinθ }} - \frac{1}{{sinθ }}} \right)}}{{\left( {\frac{1}{{cosθ }} + \frac{{sinθ }}{{cosθ }}\;} \right)\left( {1 - sinθ } \right)}}\)

\( = \;\frac{{\left( {\frac{1}{{sinθ cosθ }}} \right)\left( {cosθ + sinθ + 1} \right)\left( {sinθ + cosθ - 1} \right)}}{{\frac{1}{{cosθ }}\left( {1 + sinθ } \right)\left( {1 - sinθ } \right)}}\)

\( = \frac{1}{{sinθ }}\;\;\frac{{\left[ {{{\left( {cosθ + sinθ } \right)}^2} - {1^2}} \right]}}{{\left( {1 - {{\sin }^2}θ } \right)}}\)

\( = \frac{1}{{sinθ }}\;\frac{{\left( {{{\cos }^2}θ + {{\sin }^2}θ + 2sinθ cosθ } \right) - 1}}{{\left( {1 - {{\sin }^2}θ } \right)}}\)

\( = \frac{1}{{sinθ }}\;\frac{{2sinθ cosθ }}{{{{\cos }^2}θ }}\)

= 2/cosθ 

= 2secθ 

∴ the required value is 2secθ .

GIVEN:

\(\frac{{tan\theta \left( {1 - sin\theta } \right)\left( {sec\theta \; + \;tan\theta } \right)}}{{\left( {1\; + \;cos\theta } \right)\left( {cosec\theta - cot\theta } \right)}}\)

FORMULA USED:

1 – sin²θ = cos²θ; 1 – cos²θ = sin²θ

(a + b)(a – b) = a² – b²

CALCULATION:

\(\frac{{tan\theta \left( {1 - sin\theta } \right)\left( {sec\theta \; + \;tan\theta } \right)}}{{\left( {1\; + \;cos\theta } \right)\left( {cosec\theta - cot\theta } \right)}}\)

⇒ \(\frac{{tan\theta \left( {1 - sin\theta } \right)\left( {\frac{1}{{cos\theta }}\; + \;\frac{{sin\theta }}{{cos\theta }}} \right)}}{{\left( {1\; + \;cos\theta } \right)\left( {\frac{1}{{sin\theta }}\; - \;\frac{{cos\theta }}{{sin\theta }}} \right)}}\)

⇒ \(\frac{{tan\theta \left( {1 - sin\theta } \right)\left( {1\; + \;sin\theta } \right)sin\theta }}{{\left( {1\; + \;cos\theta } \right)\left( {1 - cos\theta } \right)cos\theta }}\)

⇒ \(\frac{{tan\theta \left( {1 - {{\sin }^2}\theta } \right)}}{{\left( {1 - {{\cos }^2}\theta } \right)}}\; \times \;tan\theta \)

⇒ \(\frac{{tan\theta \left( {{{\cos }^2}\theta } \right)}}{{{{\sin }^2}\theta }}\; \times \;tan\theta \)

⇒ tan²θ × cot²θ

⇒ 1

GIVEN:

cot²θ[(cosec θ – cot θ)(tan θ + sin θ)]sec θ

FORMULA USED:

1 – cos²x = sin²x

1 – sin²x = cos²x

CALCULATION:

cot²θ[(cosec θ – cot θ)(tan θ + sin θ)]sec θ

⇒ cot²θ[(1/sin θ – cos θ/sin θ)(sin θ/cos θ + sin θ)]sec θ

⇒ cot²θ[(1 – cos θ)(1 + cos θ)]sec θ/(cos θ)

⇒ cot²θ[1 – cos²θ]sec θ/(cos θ)

⇒ cot²θ(sin²θ)/(cos²θ)

⇒ cot²θ × tan²θ

⇒ 1

Concept

Cot x = Base/Perpendicular

Sin x = Perpendicular/Hypotenuse

Cos x = Base/Hypotenuse

Calculation

AC = √(4² + 3²)

⇒ √(16 + 9) = √25 = 5 units

Sin x = 3/5

Cos x = 4/5

So, 3/5 + 4/5

⇒ 7/5

Given:

The given logarithmic expression is log cot 9° + log cot 45° + log cot 81°?

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

log m + log n = log (m × n)

tan θ = cot (90 - θ)

tan θ × cot θ = 1

cot 45° = 1

log 1 = 0

Calculation:

By applying the identity of log

∴ log cot 9° + log cot 45° + log cot 81° = log ( cot 9° × cot 45° × cot 81°)

⇒ log ( cot 9° × cot 45° × cot 81°) = log ( cot 9° × cot 45° × cot (90 - 9)°)

⇒ log ( cot 9° × cot 45° × tan 9°) = log 1 = 0

Hence, option (1) is correct

Given:

2x = sinθ

2/x = cosθ

Formula used:

Sin2θ + Cos2θ = 1

Calculation:

2x = sinθ

⇒ 4x² = sin²θ

2/x = cosθ

⇒ 4/x² = cos²θ

Sin2θ + Cos2θ = 4x2 + 4/x2 

⇒ 1 = 4x2 + 4/x2 

⇒ 1 = 4(x2 + 1/x2)

∴ Required value is 1

Given:

X = tan40°

Formula used:

tan (90° – θ) = cot θ 

cot θ = 1/tan θ

Calculation:

2tan50º

⇒ 2tan(90° – 40°)

⇒ 2 × cot 40° 

⇒ 2 × 1/tan 40° 

⇒ 2 × (1/X) = 2/X

∴ The value of 2tan50º is 2/X.

Given:

The Sum of the angles = 157.5°

The difference of the angles = π/8

Concept used: 

Radian measure = (π/180) × degree measure 

Calculation:

Let us take the two angles as 'a' and 'b'

Sum of the angles = a + b = 157.5°       ----(i)

Difference between angles = a - b = π/8

π/8 = (π/180) × degree measure 

⇒ Degree measure = π/8 × (180/π) = 22.5° 

a - b = 22.5°       ----(ii)

Adding the equation (i) and (ii), we get

(a + b) + (a - b) = 157.5° +  22.5° = 180° 

⇒ 2a = 180° 

⇒ a = 90° 

Now, b = 157.5° - 90° = 67.5° 

∴ The value of greater angle is 90° 

Given:

tanθ + cotθ = 2 

θ is an acute angle

Concept used: 

radian measure = (π/180) × degree measure 

Calculation:

tanθ + cot θ = 2

⇒ tanθ + (1/tanθ ) = 2

Taking LCM,  

tan²θ + 1 = 2tanθ 

⇒ tan2θ + 1 - 2tanθ = 0

⇒ (tanθ - 1)² = 0 

Taking root on both sides 

⇒ tanθ - 1 = 0

⇒ tanθ = 1

⇒ θ = 45° 

Now converting degree to radian

Radian measure = 45 × (π/180) ⇒ π/4

∴ The value of θ in circular measure is π/4.

Calculation: 

cot(25° - θ) - sec(35° - θ) - tan(65° + θ) + cosec(55° + θ)

⇒ cot(25° - θ) - sec(35° - θ) - tan[90° - (25° - θ)] + cosec[90° - (35° - θ)]

⇒ cot(25° - θ) - sec(35° - θ) - cot(25° - θ) + sec(35° - θ)

⇒ 0

∴ The required value is 0.

Given-

sinx = 2/3

Formula Used- 

cosx = √(1 - sin²x)

cos3x = 4cos³x - 3cosx  

Calculation - 

cosx = √(1 - 4/9)

⇒ √(5/9)

⇒ √5/3

cos3x = 4 × (√5/3)³ - 3 × √5/3

⇒ cos3x = 20√5/27 - √5

⇒ cos3x = -7√5/27

∴ cos3x = -0.5797

Given:

tan θ + cot θ = 6

Formula Used:

(a + b)² = a² + b² + 2ab

tanθ × cotθ = 1

Calculations:

tan θ + cot θ = 6

Squaring both sides, we get

(tan θ + cot θ)² = (6)²

⇒ tan2 θ + cot2 θ + 2 × tanθ × cotθ = 36

⇒ tan2 θ + cot2 θ + 2 × 1 = 36

⇒ tan2 θ + cot2 θ = 36 – 2

⇒ tan2 θ + cot2 θ = 34

∴ The value of tan2 θ + cot2 θ is 34.

Given - 

sin θ + sin2 θ = 1

Formula used - 

sin²θ + cos²θ = 1

Solution - 

sin θ + sin2 θ = 1

⇒ sinθ = 1 - sin²θ = cos²θ

⇒ sinθ = cos²θ 

⇒ cos2θ + cos4θ = sinθ + sin²θ = 1

∴ cos2θ + cos4θ = 1.

Concept:

1 + tan2 θ = sec2 θ

sec2 θ - 1 = tan2 θ

Calculation:

Given: sec2 θ + tan2 θ = 3

To Find: Value of sec θ

sec2 θ + tan2 θ = 3

Subtracting 1 both sides, we get

⇒ sec2 θ + tan2 θ - 1 = 3 - 1

⇒ tan2 θ + tan2 θ = 2              (∵ sec2 θ - 1 = tan2 θ)

⇒ 2tan2 θ = 2

⇒ tan2 θ = 1

∴ tan θ = 1

Now,

cot θ = \(\rm \frac {1}{\tan \theta} = \frac 1 1 = 1\)

Given:

(sin θ + cosec θ)² + (cos θ +  sec θ)² = k + tan² θ + cot² θ 

Formula used:

(a + b)² = a² + 2ab + b²

sin² θ + cos² θ = 1

sec² θ - tan² θ = 1

cosec² θ - cot² θ = 1

sin θ × cosec θ = 1

cos θ × sec θ = 1

Calculation:

According to the question,

sin² θ + cosec² θ + 2 × sin θ × cosec θ + cos² θ + sec² θ + 2 × cos θ × sec θ = k + tan² θ + cot² θ 

⇒ 1 + cosec² θ + 2 × 1 + sec² θ + 2 × 1 = k + tan² θ + cot² θ 

⇒ 5 + (cosec² θ - cot² θ) + (sec² θ - cot² θ) = k

⇒ 5 + 1 + 1 = k

⇒ k = 7

∴ The value of k is 7.