Question

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Answer 1

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{A + B + C = 45\degree} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{ \sum \: (tanA + tantanB)}\end{cases}\end{gathered}\end{gathered}$

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

$\boxed{ \red{ \sf \: tan(x + y)\: = \dfrac{tanx + tany}{1 - tanx \: tany}}}$

$\boxed{ \red{ \sf \: tan(x - y)\: = \dfrac{tanx - tany}{1 + tanx \: tany}}}$

$\boxed{ \red{ \sf \:tan45\degree = 1}}$

Solution :-

$\rm :\longmapsto\:A + B + C = 45\degree$

$\rm :\longmapsto\:A + B = 45\degree - C$

$\rm :\longmapsto\:tan(A + B) = tan(45\degree - C)$

$\rm :\longmapsto\:\dfrac{tanA + tanB}{1 - tanA \: tanB} = \dfrac{tan45\degree - tanC}{1 + tan45\degree \: tanC}$

$\rm :\longmapsto\:\dfrac{tanA + tanB}{1 - tanA \: tanB} = \dfrac{1 - tanC}{1 + tanC}$

$\rm \: (tanA + tanB)(1 + tanC) = (1 - tanAtanB)(1 - tanC)$

$\rm \: tanA + tanB + tanAtanC + tanBtanC = \\ \rm \: 1 - tanC - tanAtanB + tanAtanBtanC \: \: \:$

$\rm \: tanA + tanB + tanC + tanAtanB + tanBtanC + \\ \rm \: tanCtanA = 1 + tanAtanBtanC \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\rm \: (tanA + tanAtanB) + (tanB + tanBtanC) + \\ \rm \:(tanC + tanCtanA) = 1 + tanAtanBtanC \: \: \: \:$

$\rm :\longmapsto\: \sum(tanA + tanAtanB) = 1 + \prod \: tanA$

$\boxed{ \red{ \bf \:Hence, \: option \: (c) \: is \: correct}}$

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)

Answer 2

${\boxed{\underline{\tt{\orange{Required \: Answer:-}}}}}$

$\displaystyle\sf\large{(c.) \: \: 1+ \pi \: \: tan A }$

★GIVEN:-

• A + B + C = 45°

★TO FIND:-

• Σ (tan A + tan A tan B ) = ??

★SOLUTION:-

As Σ ( tan A + tan A tan B ) can be expanded as

⇢ tan A + tan B + tan C + tan A tan B + tan B tan C + tan C tan A

∵ A + B + C = 45°

Taking tan on both the sides

⇝ tan ( A + B + C) = tan 45°

$\displaystyle \sf{ ⇝ \frac{tan \: A \: + \: tan(B + C)}{1 -tan(B + C) tan \: A} = 1}$

⇝ tan A + tan ( B + C ) = 1 - tan (B+C) tan A

$\displaystyle \sf{⇝tan \: A \: + \frac{tanB + \: tan C}{1 -tanB \: tan C } = 1 -\frac{tanB + \: tan C}{1 -tanB \: tan C } \: \: tan \: A}$

$\displaystyle \sf{⇝ \frac{tan \: A \: (1 - tanB \: tan C) + tanA+ \: tan C}{ \cancel{1 - tanB \: tan C}} = \frac{1 -tanB \: tan C( - tanA \: (tanB \: tan C) }{ \cancel{1 -tanB \: tan C}} }$

$\displaystyle \sf{⇝tan \: A \: -tanA \:tanB \: tan C \: + tan \: A + tan C = 1 -tanB \: tan C - tanA \:tanB - tan \: A tan C }$

$\displaystyle \sf {⇝ \: tanA + \:tanB + \: tan C + \: tan \: A \: \: tan C + tanA \:tanB +tanB \: tan C = 1 + tanA \:tanB \: tan C }$

$\displaystyle \sf {⇝ \sum \: (tan \: A + tanA \:tanB) = 1 + tan \: A \: tanB \: tan C}$

$\displaystyle \sf {⇝ \sum \: (tan \: A \: + \: tanA \:tanB) = \pink{ 1 + \pi \:tanA} }$