Question

if sin (theta + Alpha) = cos (theta + Alpha) prove that tan theta = 1 - tan Alpha/oneplus tan Alpha
$if \sin(theta + \alpha ) \: = \cos(theta + \alpha ) prove \: that \: \tan(theta) = 1 + \tan\alpha upon \: 1 - \tan\alpha$
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Answer 1

Answer:-

Given:-

sin (θ + α) = cos (θ + α)

⟹ sin (θ + α) / cos (θ + α) = 1

using tan θ = sin θ/cos θ we get,

⟹ tan (θ + α) = 1

⟹ tan (θ + α) = tan 45°

[ tan 45° = 1 ]

On comparing both sides we get;

⟹ θ + α = 45°

⟹ θ = 45° + (- α)

Again applying tan on both sides we get,

⟹ tan θ = tan [ 45° + ( - α) ]

Using tan (A + B) = (tan A + tan B) / 1 - tan A tan B in RHS we get,

$\\ \implies \sf \tan \theta = \frac{tan \: 45 ^{ \circ} + tan \: ( -\alpha)}{1 - \tan 45 ^{ \circ} \tan \: (- \alpha) } \\ \\ \\\implies \sf \tan \theta = \frac{1 - \tan \: \alpha }{1 - (1)(- \tan \: \alpha ) } \:\:\:\:\: \{ \tan ( - \theta) = - \tan \theta\} \\ \\ \\ \implies \sf \tan \theta = \frac{1 - \tan \: \alpha }{1 + \tan \: \alpha }$

Hence, Proved.

Answer 2

Gɪᴠᴇɴ :

$\red\checkmark\:\:\bf{\sin\:(\theta\:+\:\alpha)\:=\:\cos\:(\theta\:+\:\alpha)\:}$

Tᴏ Pʀᴏᴠᴇ :

• $\sf{\tan\theta\:=\:\dfrac{1\:+\:\tan\alpha}{1\:-\:\tan\alpha}}$

Pʀᴏᴏғ :

Gɪᴠᴇɴ ᴛʜᴀᴛ,

➛ $\sf{\sin\:(\theta\:+\:\alpha)\:=\:\cos\:(\theta\:+\:\alpha)\:}$

➛ $\sf{\dfrac{\sin\:(\theta\:+\:\alpha)}{\cos\:(\theta\:+\:\alpha)}\:=\:1\:}$

➛ $\sf{\tan\:(\theta\:+\:\alpha)\:=\:1\:}$

➛ $\sf{\tan\:(\theta\:+\:\alpha)\:=\:\tan{45^{\circ}}\:}$

➛ $\bf{\theta\:+\:\alpha\:=\:45^{\circ}\:}$

➛ $\sf{\theta\:=\:45^{\circ}\:-\:\alpha}$

↝ Using tan function on both side,

➛ $\sf{\tan\theta\:=\:\tan{45^{\circ}\:-\:\alpha}}$

↝ Using the below relation,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

$\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{tan\:(\theta\:-\:\beta)\:=\:\dfrac{tan\theta\:-tan\beta}{1\:+\:tan\theta\:tan\beta}\:}}}}}}$

$:\implies\:\:\sf{\tan\theta\:=\:\dfrac{\tan{45^{\circ}}\:-\:\tan\alpha}{1\:+\:\tan{45^{\circ}}\:\tan\alpha}\:}$

$:\implies\:\:\sf{\tan\theta\:=\:\dfrac{1\:-\:\tan\alpha}{1\:+\:(1\times{\tan\alpha})}\:}$

$:\implies\:\:{\underline{\boxed{\bf{\pink{\tan\theta\:=\:\dfrac{1\:-\:\tan\alpha}{1\:+\:\tan\alpha}}}}}}\:(Hence\:Proved)$