Question

Prove: tan x/2=sinx/1+cosx

Answer 1

Required Answer:-

Given To Prove:

• tan(x/2) = sin(x)/(1 + cos(x))

Proof:

To prove this question, we will prove that,

→ LHS = RHS i.e.,

→ Left Hand Side = Right Hand Side.

Taking LHS,

$\sf = \tan({}^{x}/_{2})$

$\sf = \dfrac{\sin({}^{x}/_{2})}{\cos({}^{x}/_{2})} \: \: \: \: \: \: \bigg( \rm as \tan(x) = \dfrac{ \sin(x) }{ \cos(x) } \bigg)$

Multiplying the numerator and denominator by 2cos(x/2), we get,

$\sf = \dfrac{2\sin({}^{x}/_{2}) \cos({}^{x}/_{2}) }{2\cos^{2} ({}^{x}/_{2})}$

Adding 1 and subtracting 1 in the denominator part, we get,

$\sf = \dfrac{2\sin({}^{x}/_{2}) \cos({}^{x}/_{2}) }{2\cos^{2} ({}^{x}/_{2}) - 1 + 1}$

Now, we know that,

$\sf \mapsto \cos(2x) = 2 \cos^{2} (x) - 1$

So,

$\sf \dfrac{2\sin({}^{x}/_{2}) \cos({}^{x}/_{2}) }{2\cos^{2} ({}^{x}/_{2}) - 1 + 1}$

$\sf = \dfrac{2\sin({}^{x}/_{2}) \cos({}^{x}/_{2}) }{(2\cos^{2} ({}^{x}/_{2}) - 1 )+ 1}$

$\sf = \dfrac{2\sin({}^{x}/_{2}) \cos({}^{x}/_{2}) }{\cos(2 \times {}^{x}/_{2}) + 1}$

$\sf = \dfrac{2\sin({}^{x}/_{2}) \cos({}^{x}/_{2}) }{\cos(x) + 1}$

Now, we know that,

$\sf \mapsto \sin(2x) = 2 \sin(x) \cos(x)$

So,

$\sf\dfrac{2\sin({}^{x}/_{2}) \cos({}^{x}/_{2}) }{\cos(x) + 1}$

$\sf = \dfrac{\sin(2 \times {}^{x}/_{2})}{\cos(x) + 1}$

$\sf = \dfrac{\sin(x)}{\cos(x) + 1}$

$\sf = \dfrac{\sin(x)}{1 + \cos(x) }$

= RHS (Hence Proved)

Therefore, we can see that,

$\sf \implies\tan({}^{x}/_{2}) = \dfrac{ \sin(x) }{1 + \cos(x) }$

•••♪

Answer 2

$\large\underline\blue{\bold{To \: Prove - }}$

$\tt \: tan\dfrac{x}{2} = \dfrac{sinx}{1 + cosx}$

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

$(1). \: \:\boxed{\red{\bf\:sin2x = 2sinx \: cosx }}$

$(2). \: \:\boxed{\red{\bf\: 1 + cos2x = 2 {cos}^{2} x}}$

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

$\bf \: Consider \: RHS$

$\rm :\implies\:\dfrac{sinx}{1 + cosx}$

$\rm :\implies\:\dfrac{2 \: sin\dfrac{x}{2} \: cos\dfrac{x}{2} }{2 {cos}^{2} \dfrac{x}{2} }$

$\rm :\implies\:\dfrac{sin\dfrac{x}{2} }{cos\dfrac{x}{2} }$

$\rm :\implies\:tan\dfrac{x}{2}$

$\rm :\implies\: = \: LHS$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information -

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)