Question

If tan A and tan B are the roots of x^2 – ax + b = 0, then the value of cos^2(A + B) is :-

1. a^2 / (a^2 + (1-b)^2
2.a^2 / (a^2 + (1+b)^2
3. 1+b / (a^2 + (1-b)^2
4. a^2 / (a+b)^2

Answer 1

EXPLANATION.

tan(A) and tan(B) are the roots of the equation.

⇒ x² - ax + b = 0.

As we know that,

Sum of the zeroes of the quadratic polynomial.

⇒ α + β = - b/a.

⇒ tan(A) + tan(B) = -(-a)/1 = a.

⇒ tan(A) + tan(B) = a. - - - - - (1).

Products of the zeroes of the quadratic polynomial.

⇒ αβ = c/a.

⇒ tan(A) x tan(B) = b. - - - - - (2).

As we know that,

Formula of :

⇒ tan(A + B) = [tan(A) + tan(B)]/[1 - tan(A) x tan(B)].

⇒ tan(A + B) = [a]/[1 - b].

Squaring on both sides of the equation, we get.

⇒ tan²(A + B) = (a)²/1 - b)².

As we know that,

Formula of :

⇒ sec²x = 1 + tan²x.

Put the values in the equation, we get.

⇒ sec²(A + B) = 1 + (a)²/(1 - b)².

⇒ sec²(A + B) = [(1 - b)² + (a)²/(1 - b)²].

⇒ 1/cos²(A + B) = [(1 - b)² + (a)²/(1 - b)²].

⇒ cos²(A + B) = [(1 - b)²/(1 - b)² + (a)²].

Answer 2

⠀⠀⠀⠀⠀★ G I V E N ⠀P O L Y N O M I A L :⠀ x² - ax + b = 0

⠀▪︎⠀By Comparing Given Polynomial with Standard form of Quadratic Equation ( i.e. ax² + bx + c = 0 ) we get —

• a = 1 ,
• b = – a &
• c = b

$\dag\:\underline {\frak{ As\:We\:Know\:that \:\::\:}}$

$\qquad \maltese \:\underline {\pmb{\sf Sum \:of \:zeroes \:\:(\:\alpha\:+\:\beta\:)\:}}$

$\dashrightarrow \sf \:\Big\{ \alpha \:+\:\beta \:\Big\}\:=\:\dfrac{-\:b\:}{a}\:\\\\\\ \dashrightarrow \sf \:\Big\{ Tan\:A \:+\:Tan\:B \:\Big\}\:=\:\dfrac{-\:(\:-a\:)\:}{1}\:\\\\\\ \dashrightarrow \sf \: Tan\:A \:+\:Tan\:B \:=\:\dfrac{a\:}{1}\:\\\\\\\dashrightarrow \pmb{\sf \: Tan\:A \:+\:Tan\:B \:=\:a\:}\:$

$\qquad \maltese \:\underline {\pmb{\sf Product \:of \:zeroes \:\:(\:\alpha\:\:\beta\:)\:}}$

$\dashrightarrow \sf \:\Big\{ \alpha \:\beta \:\Big\}\:=\:\dfrac{c\:}{a}\:\\\\\\ \dashrightarrow \sf \:\Big\{ Tan\:A \:Tan\:B \:\Big\}\:=\:\dfrac{b\:}{1}\:\\\\\\ \dashrightarrow \sf \: Tan\:A \:Tan\:B \:=\:b\:\\\\\\\dashrightarrow \pmb{\sf \: Tan\:A \:Tan\:B \:=\:b\:}\:$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\\\:\bigstar \:\underline{\sf According \:To\:The\:Question \:\::\:}$

$:\implies \sf \: Tan \:( \: A + B\: )\:=\: \dfrac{ Tan\:A \:+\:Tan\:B }{ 1 - Tan\:A \times Tan \:B \:}\\\\\\:\implies \sf \: Tan \:( \: A + B\: )\:=\: \dfrac{ a }{ 1 - b \:}\\\\\\:\implies \sf \:\Big\{ \: Tan \:( \: A + B\: )\:\Big\}^2\:=\:\bigg\{\: \dfrac{ a }{ 1 - b \:}\:\bigg\}^2\\\\\\:\implies \sf \: \: Tan^2 \:( \: A + B\: )\:\:=\: \bigg\{ \dfrac{ a^2 }{ (\:1 - b \:)^2\:} \bigg\}\:$

• Sec² x = 1 + Tan² x

$\\:\implies \sf Sec^2 \:( \: A + B\: )\:\:=\: 1 \:+\:Tan^2\:(\:A + B\:)\:\\\\\\\\:\implies \sf Sec^2 \:( \: A + B\: )\:\:=\:\Bigg[\: 1 \:+\:\dfrac{ a^2 }{ (\:1 - b \:)^2\:} \:\Bigg]\:\\\\\\\\:\implies \sf Sec^2 \:( \: A + B\: )\:\:=\: \:\Bigg[ \dfrac{\: (\:1 - b \:)^2 + a^2 }{ \: (\:1 - b \:)^2\:}\:\Bigg] \:\\\\\\\\:\implies \sf \dfrac{1}{Cos^2} \:( \: A + B\: )\:\:=\: \:\Bigg[ \dfrac{\: (\:1 - b \:)^2 + a^2 }{ \: (\:1 - b \:)^2\:} \Bigg]\:\qquad \:\:\:\:\quad \because \red{ \:\Bigg\lgroup Sec\: A \:=\:\dfrac{1}{Cos\:A}\:\Bigg\rgroup } \\\\\\\\:\implies \sf \dfrac{1}{Cos^2} \:( \: A + B\: )\:\:=\: \:\:\Bigg[ \dfrac{\: (\:1 - b \:)^2 + a^2 }{ \: (\:1 - b \:)^2\:} \Bigg] \\\\\\\\:\implies \:\underline{\boxed{\purple{\frak{ Cos^2 \:( \: A + B\: )\:\:=\: \:\Bigg[ \dfrac{(\:1 - b\:)^2 }{\: (\:1 - b \:)^2 + a^2 }\Bigg]}}}}\:\:\bigstar$

$\therefore \:\sf Hence ,\:Value \:of \:Cos^2\:(\:A + B\:) \:is \:\red{ \Bigg[ \dfrac{(\:1 - b\:)^2} {\: (\:1 - b \:)^2 + a^2 }\:\Bigg]}\:.$