Multiple and Sub-multiple Angles MCQ Quiz - Objective Question with Answer for Multiple and Sub-multiple Angles - Download Free PDF

Given:

In a triangle, the relationship between angles is used where A + B + C = 180°.

Formula used:

$\cos \left(\frac{A+B}{2}\right)=\sin \left(\frac{C}{2}\right)$, where C = 180° - (A + B).

Explanation:

In a triangle:

A + B + C = 180°

⇒ C = 180° - (A + B)

⇒ $\frac{C}{2}=\frac{180°-(A+B)}{2}$

Now, using the complementary angle relationship:

$\cos \left(\frac{A+B}{2}\right)=\sin \left(\frac{C}{2}\right)$

∴ Option 4 is the correct answer.

Given:

In a triangle, the relationship between angles is used where A + B + C = 180°.

Formula used:

$\cos \left(\frac{A+B}{2}\right)=\sin \left(\frac{C}{2}\right)$, where C = 180° - (A + B).

Explanation:

In a triangle:

A + B + C = 180°

⇒ C = 180° - (A + B)

⇒ $\frac{C}{2}=\frac{180°-(A+B)}{2}$

Now, using the complementary angle relationship:

$\cos \left(\frac{A+B}{2}\right)=\sin \left(\frac{C}{2}\right)$

∴ Option 4 is the correct answer.

$⟹2b=a+c$

The value of the semiperimeter of the triangle $s={\displaystyle \frac{3b}{2}}$

$\therefore$ the value of ${\cos }^2{\displaystyle \frac{A}{2}}={\displaystyle \frac{s(s-a)}{bc}}={\displaystyle \frac{3({\displaystyle \frac{3b}{2}}-a)}{2c}}$

Similarly the value of ${\cos }^2{\displaystyle \frac{C}{2}}={\displaystyle \frac{s(s-c)}{ab}}={\displaystyle \frac{3({\displaystyle \frac{3b}{2}}-c)}{2a}}$

$a{\cos }^2{\displaystyle \frac{C}{2}}+c{\cos }^2{\displaystyle \frac{A}{2}}={\displaystyle \frac{3}{2}}(3b-a-c)={\displaystyle \frac{3}{2}}\left(b\right)$

Calculation

${(a-b)}^2{\cos }^2{\displaystyle \frac{C}{2}}+{(a+b)}^2{\sin }^2{\displaystyle \frac{C}{2}}=a^2+b^2+2ab({\sin }^2{\displaystyle \frac{C}{2}}-{\cos }^2{\displaystyle \frac{C}{2}})=a^2+b^2+2ab(2{\sin }^2{\displaystyle \frac{C}{2}}-1)$

But $\sin {\displaystyle \frac{C}{2}}=\sqrt{{\displaystyle \frac{(s-a)(s-b)}{ab}}}$

where $s={\displaystyle \frac{a+b+c}{2}}$

$\therefore$ $a^2+b^2+2ab(2{\sin }^2{\displaystyle \frac{C}{2}}-1)=a^2+b^2+2ab({\displaystyle \frac{(c+b-a)(c+a-b)}{2ab}}-1)=a^2+b^2+c^2-{(b-a)}^2-2ab=c^2$

Hence option 2 is correct

Concept Used:

Trigonometric identities:

sin C + sin D = 2 sin $\frac{C+D}{2}$ cos $\frac{C-D}{2}$

cos C + cos D = 2 cos $\frac{C+D}{2}$ cos $\frac{C-D}{2}$

tan θ = $\frac{sin\theta }{cos\theta }$

Calculation:

⇒ $\frac{sin7x+sin5x}{cos7x+cos5x}$

⇒ $\frac{2sin\frac{7x+5x}{2}cos\frac{7x-5x}{2}}{2cos\frac{7x+5x}{2}cos\frac{7x-5x}{2}}$

⇒ $\frac{2sin6xcosx}{2cos6xcosx}$

⇒ $\frac{sin6x}{cos6x}$

⇒ tan 6x

∴ $\frac{sin7x+sin5x}{cos7x+cos5x}$ = tan 6x

Hence option 5 is correct

Concept:

cos 2A = cos² A – sin² A

sin 2A = 2 sin A × cos A

Calculation:

$\Rightarrow \cot \left(\frac{A}{2}\right)-\tan \left(\frac{A}{2}\right)=\frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}-\frac{\sin \left(\frac{A}{2}\right)}{\cos \left(\frac{A}{2}\right)}=\frac{{\cos }^2\left(\frac{A}{2}\right)-{\sin }^2\left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)\times \cos \left(\frac{A}{2}\right)}$

As we know that, cos 2A = cos² A – sin² A

$\Rightarrow \cot \left(\frac{A}{2}\right)-\tan \left(\frac{A}{2}\right)=\frac{{\cos }^2\left(\frac{A}{2}\right)-{\sin }^2\left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)\times \cos \left(\frac{A}{2}\right)}=\frac{\cos A}{\frac{1}{2}\times 2\times \sin \left(\frac{A}{2}\right)\times \cos \left(\frac{A}{2}\right)}$

As we know that, sin 2A = 2 sin A × cos A

$\Rightarrow \cot \left(\frac{A}{2}\right)-\tan \left(\frac{A}{2}\right)=\frac{\cos A}{\frac{1}{2}\times 2\times \sin \left(\frac{A}{2}\right)\times \cos \left(\frac{A}{2}\right)}=2\frac{\cos A}{\sin A}=2\cot A$

Concept:

cos 2A = cos² A - sin² A = 2 cos² A - 1

sin 2A = 2 sin A × cos A

Calculation:

$\Rightarrow \cot A+cosecA=\frac{\cos A}{\sin A}+\frac{1}{\sin A}$

$\Rightarrow \cot A+cosecA=\frac{1+\cos A}{\sin A}$

As we know that,

cos 2A = cos² A - sin² A = 2 cos² A - 1

$\Rightarrow \cot A+cosecA=\frac{1+\cos A}{\sin A}=\frac{2\times {\cos }^2\left(\frac{A}{2}\right)}{\sin A}$

As we know that,

sin 2A = 2 sin A × cos A

$\Rightarrow \cot A+cosecA=\frac{1+\cos A}{\sin A}$

$=\frac{2\times {\cos }^2\left(\frac{A}{2}\right)}{\sin A}=\frac{2\times {\cos }^2\left(\frac{A}{2}\right)}{2\times \cos \left(\frac{A}{2}\right)\times \sin \left(\frac{A}{2}\right)}=\cot \left(\frac{A}{2}\right)$

∴ The value of cot A + cosec A is $\cot \left(\frac{A}{2}\right)$

Concept:

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

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

Calculation:

sin 75°

= sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30° 

= $\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}$

= $\frac{\sqrt{3}+1}{2\sqrt{2}}$

Concept:

sin² x + cos² x = 1

sin 2x = 2 sin x cos x

Calculation:

Given: sin α + cos α = p

By squaring both the sides, we get

⇒ sin² α + cos² α + 2 sin α cos α = p²

As we know that, sin² x + cos² x = 1 and sin 2x = 2 sin x cos x

⇒ 1 + sin 2α = p²

⇒ sin 2α = p² – 1

As we can write cos² 2α = 1 – sin² 2α

⇒ cos2 2α = 1 – (p² – 1)²

Concept:

• cos 2A = cos² A – sin² A = 1 – 2 sin² A = 2 cos² A – 1
• $\sin \frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{2}}$
• $\cos \frac{A}{2}=\pm \sqrt{\frac{1+\cos A}{2}}$

Calculation:

As we know that, cot x = cos x / sin x

$\Rightarrow \cot \left(22{\frac{1}{2}}^{\circ }\right)=\frac{\mathrm{c}\mathrm{o}\mathrm{s}\left(22{\frac{1}{2}}^{\circ }\right)}{\sin \left(22{\frac{1}{2}}^{\circ }\right)}$

As we know that, $\sin \frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{2}}$

$\Rightarrow \sin \left({\frac{45}{2}}^{\circ }\right)=\pm \sqrt{\frac{1-\cos \left({45}^{\circ }\right)}{2}}$

As we know that, 0° < A / 2 < 90° where all trigonometric ratios are positive.

$\Rightarrow \sin \left({\frac{45}{2}}^{\circ }\right)=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}$      -------(1)

As we know that, $\cos \frac{A}{2}=\pm \sqrt{\frac{1+\cos A}{2}}$

$\Rightarrow \cos \left({\frac{45}{2}}^{\circ }\right)=\pm \sqrt{\frac{1+\cos \left({45}^{\circ }\right)}{2}}$      -------(2)

As we know that, 0° < A / 2 < 90° where all trigonometric ratio’s are positive.

$\Rightarrow \cos \left({\frac{45}{2}}^{\circ }\right)=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}$

So, from equation (1) and (2), we get

$\Rightarrow \cot \left(22{\frac{1}{2}}^{\circ }\right)=\frac{\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}}{\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}}=1+\sqrt{2}$

$\Rightarrow \sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}$

Rationalization above the equation

$\Rightarrow \sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}}$

$\Rightarrow \sqrt{\frac{(\sqrt{2}+1{)}^2}{(\sqrt{2}{)}^2-(1{)}^2}}$       (∵ a² - b² = (a + b)(a - b)

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

CONCEPT:

Trigonometric Identities:

• $\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}\theta =\frac{1}{\mathrm{s}\mathrm{i}\mathrm{n}\theta }$
• $\mathrm{s}\mathrm{e}\mathrm{c}\theta =\frac{1}{\mathrm{c}\mathrm{o}\mathrm{s}\theta }$
• $\mathrm{c}\mathrm{o}\mathrm{s}{30}^{\circ }=\mathrm{s}\mathrm{i}\mathrm{n}{60}^{\circ }=\frac{\sqrt{3}}{2}$
• $\mathrm{c}\mathrm{o}\mathrm{s}{60}^{\circ }=\mathrm{s}\mathrm{i}\mathrm{n}{30}^{\circ }=\frac{1}{2}$
• (cos A× cos B) – (sin A × sin B) = cos(A + B)
• 2 × sin A × cos A = sin (2A)
• sin (90 - θ) = cos θ

CALCULATION:

Given that: √3 cosec 20° – sec 20°

As we know that, cosec θ = 1/ sin θ and sec θ = 1/ cos θ

$\Rightarrow \frac{\sqrt{3}}{\sin {20}^{\circ }}-\frac{1}{\cos {20}^{\circ }}$

$\Rightarrow 2\left\{\frac{\frac{\sqrt{3}}{2}}{\sin {20}^{\circ }}-\frac{\frac{1}{2}}{\cos {20}^{\circ }}\right\}$

As we know that, cos 30° = √3 / 2, sin 30° = ½

$\Rightarrow 2\left\{\frac{\cos {30}^{\circ }}{\sin {20}^{\circ }}-\frac{\sin {30}^{\circ }}{\cos {20}^{\circ }}\right\}$

$\Rightarrow 2\left\{\frac{\cos {30}^{\circ }\cos {20}^{\circ }-\sin {30}^{\circ }\sin {20}^{\circ }}{\sin {20}^{\circ }\cos {20}^{\circ }}\right\}$

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

$\Rightarrow 2\left\{\frac{\cos \left({50}^{\circ }\right)}{\sin {20}^{\circ }\cos {20}^{\circ }}\right\}$

$\Rightarrow 2\times 2\left\{\frac{\cos \left({50}^{\circ }\right)}{2\times \sin {20}^{\circ }\cos {20}^{\circ }}\right\}$

As we know that, 2 × sin A × cos A = sin (2A)

$\Rightarrow 4\left\{\frac{\cos \left({50}^{\circ }\right)}{\sin {40}^{\circ }}\right\}$

$\Rightarrow 4\left\{\frac{\cos \left({50}^{\circ }\right)}{\mathrm{s}\mathrm{i}\mathrm{n}\left({90}^{\circ }-{50}^{\circ }\right)}\right\}$

As we know that, sin (90 - θ) = cos θ

$\Rightarrow 4\left\{\frac{\cos \left({50}^{\circ }\right)}{\cos {50}^{\circ }}\right\}$

Concept:

a² - b² = (a + b)(a - b)

sin C + sin D = $2\sin (\frac{\mathrm{C}+\mathrm{D}}{2})\cos (\frac{\mathrm{C}-\mathrm{D}}{2})$

sin C - sin D = $2\cos (\frac{\mathrm{C}+\mathrm{D}}{2})\sin (\frac{\mathrm{C}-\mathrm{D}}{2})$

2 sin x cos x = sin 2x

Calculation:

Consider, sin2 6x – sin2 4x

= (sin 6x + sin 4x) (sin 6x – sin 4x)

= [ $2\sin (\frac{6\mathrm{x}+4\mathrm{x}}{2})\cos (\frac{6\mathrm{x}-4\mathrm{x}}{2})$ ] [$2\cos (\frac{6\mathrm{x}+4\mathrm{x}}{2})\sin (\frac{6\mathrm{x}-4\mathrm{x}}{2})$]

= $[2\sin 5\mathrm{x}\cos \mathrm{x}][2\cos 5\mathrm{x}\sin \mathrm{x}]$

Rearranging the terms, we get

= $[2\sin 5\mathrm{x}\cos 5\mathrm{x}][2\sin \mathrm{x}\cos \mathrm{x}]$

= sin 10x sin 2x

= sin 2x sin 10x

Concept:

cosec (2nπ - θ) = - cosec θ

Calculation:

cosec (-1410°) = -cosec (1410°)

= -cosec (4× 360° - 30°)

Since cosec(2nπ - θ) = - cosecθ, we can write:

- cosec (4×360° - 30°) = - (-cosec (30°))

= cosec 30° 

= 2

Hence, option (3) is correct.

Concept:

cos (90 - x) = sin x

sin A - sin B = 2cos$\frac{(\mathrm{A}+\mathrm{B})}{2}$sin $\frac{(\mathrm{A}-\mathrm{B})}{2}$

cos 45° = $\frac{1}{\sqrt{2}}$

Calculation:

Given,

cos18° - sin18°

= cos (90° - 72°) - sin 18°

= sin 72° - sin 18°

= 2 cos $\frac{(72°+18°)}{2}$× sin $\frac{(72°-18°)}{2}$

= 2 cos 45° sin 27°

= 2× $\frac{1}{\sqrt{2}}$× sin 27°

= √2 sin 27° 

Concept:

$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}=2\mathrm{c}\mathrm{o}\mathrm{s}\frac{\mathrm{A}+\mathrm{B}}{2}\mathrm{s}\mathrm{i}\mathrm{n}\frac{\mathrm{A}-\mathrm{B}}{2}$

$\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{A}+\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{B}=2\mathrm{c}\mathrm{o}\mathrm{s}\frac{\mathrm{A}+\mathrm{B}}{2}\mathrm{c}\mathrm{o}\mathrm{s}\frac{\mathrm{A}-\mathrm{B}}{2}$

Calculation:

$\frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{A}+\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{B}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}=$

$\frac{2\mathrm{c}\mathrm{o}\mathrm{s}\frac{\mathrm{A}+\mathrm{B}}{2}\mathrm{c}\mathrm{o}\mathrm{s}\frac{\mathrm{A}-\mathrm{B}}{2}}{2\mathrm{c}\mathrm{o}\mathrm{s}\frac{\mathrm{A}+\mathrm{B}}{2}\mathrm{s}\mathrm{i}\mathrm{n}\frac{\mathrm{A}-\mathrm{B}}{2}}$

=$\mathrm{c}\mathrm{o}\mathrm{t}\frac{\mathrm{A}-\mathrm{B}}{2}$

Hence, option (4) is correct.