Question

Prove that
2Cosec 2x+Cosec x= sec x * cot x/2

Answer 1

☆ Prove that

$\tt \ \: :  ⟼ 2 \: cosec2x + cosecx = secx \: cot \: \dfrac{x}{2}$

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\tt \ \: :  ⟼ (1) \: \boxed{sin2x = \: 2sinx \: cosx}$

$\tt \ \: :  ⟼ (2) \: \boxed{ 1 + cos2x = 2 {cos}^{2} \: x}$

$\tt \ \: :  ⟼ (3) \: \boxed{ \dfrac{cosx}{sinx} \: = cot \: x}$

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$\large\underline\purple{\bold{Solution :- }}$

$\bf \:  ⟼ Consider \: LHS$

$\tt \ \: :  ⟼ 2 \: cosec2x + cosecx$

$\tt \ \: :  ⟼ \dfrac{2}{sin2x} + \dfrac{1}{sinx}$

$\tt \ \: :  ⟼ \dfrac{2}{2 \: sinx \: cosx} + \dfrac{1}{sinx}$

$\tt \ \: :  ⟼ \dfrac{1}{sinx \: cosx} + \dfrac{1}{sinx}$

$\tt \ \: :  ⟼ \dfrac{1 + cosx}{sinx \: cosx}$

$\tt \ \: :  ⟼ \dfrac{2 \: {cos}^{2} \dfrac{x}{2} }{2 \: sin\dfrac{x}{2} \: cos\dfrac{x}{2} \: cosx}$

$\tt \ \: :  ⟼ \dfrac{cos\dfrac{x}{2}}{sin\dfrac{x}{2} \times cosx}$

$\tt \ \: :  ⟼ secx \times cot\dfrac{x}{2}$

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Additional Information

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

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

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

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

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

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)

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Answer 2

Answer:

Given : 2\csc 2x+\csc x =cot\frac{x}{2}\sec x2csc2x+cscx=cot

2

x

secx

Step-by-step explanation:

\begin{gathered}2\csc 2x+\csc x \\\\2(\frac{1}{\sin 2x})+\frac{1}{\sin x}\\\\2(\frac{1}{2\sin x\cos x})+\frac{1}{\sin x}\\\\\frac{1}{\sin x\cos x}+\frac{1}{\sin x}\\\\\text{taking LCM}\\\\\frac{1+\cos x }{\sin x\cos x}\\\\\frac{1+\cos x }{\sin x}\times \frac{1}{\cos x}\\\\\frac{1+\cos x }{\sin x}\times \sec x\\\\\frac{2\cos^2\frac{x}{2}}{2\sin^2\frac{x}{2}.\cos^2\frac{x}{2}}\times \sec x\\\\\cot\frac{x}{2}\sec x\end{gathered}

2csc2x+cscx

2(

sin2x

1

)+

sinx

1

2(

2sinxcosx

1

)+

sinx

1

sinxcosx

1

+

sinx

1

taking LCM

sinxcosx

1+cosx

sinx

1+cosx

×

cosx

1

sinx

1+cosx

×secx

2sin

2

2

x

.cos

2

2

x

2cos

2

2

x

×secx

cot

2

x

secx