Question

Solve:
(1+tanA +secA)( 1+cotA-cosecA)​

Answer 1

$\sf\huge\red{\underbrace{ Question : }}$

Solve : (1 + tan A + sec A)(1 + cot A - cosec A)

$\sf\huge\pink{\underbrace{ Solution : }}$

$\sf \implies ( 1 + \tan\:A + \sec\:A)(1 + \cot\:A-\csc\:A)$

• tan A = sin A/cos A
• sec A = 1/cos A
• cot A = cos A/sin A
• csc A = 1/sin A

$\sf \implies \Bigg(1+\cfrac{\sin\:A}{\cos\:A} + \cfrac{1}{\cos\:A}\Bigg)\Bigg(1+\cfrac{\cos\:A}{\sin\:A} - \cfrac{1}{\sin\:A}) \Bigg)$

$\sf \implies \Bigg( \cfrac{ \cos\:A + \sin\:A + 1}{\cos\:A}\Bigg) \Bigg( \cfrac{\sin\:A + \cos\:A - 1}{\sin\:A}\Bigg)$

$\sf \implies \Bigg( \cfrac{(\cos\:A + \sin\:A)^{2} - (1)^{2}}{\cos\:A\sin\:A} \Bigg)$

• (a + b)(a - b) = a² - b²

$\sf \implies \Bigg( \cfrac{\cos^{2}\:A + \sin^{2}\:A + 2\cos\:A\sin\:A - 1}{\cos\:A\sin\:A} \Bigg)$

• (a + b)² = a² + b² + 2ab
• sin² A + cos² A = 1

$\sf \implies \Bigg( \cfrac{1 + 2\sin\:A\cos\:A - 1}{\cos\:A\sin\:A} \Bigg)$

$\sf \implies \Bigg( \cfrac{2 \: \: \cancel{\sin\:A} \: \: \cancel{\cos\:A}}{\cancel{\cos\:A} \: \: \cancel{sin\:A}}\Bigg)$

$\sf \implies 2$

$\underline{\boxed{\rm{\purple{\therefore (1+\tan\:A+\sec\:A)(1+\cot\:A-\csc\:A)=2.}}}}\:\orange{\bigstar}$

★ More Information :

$\boxed{\begin{minipage}{7 cm} Trigonometric Identities : \\ \\ :\implies\sin^{2}\theta + cos^{2}\theta = 1 \\ \\ :\implies 1 + tan^{2}\theta = sec^{2}\theta \\ \\ :\implies1 + cot^{2}\theta=\text{cosec}^2\, \theta \end{minipage}}$

______________________________________________________

Answer 2

$\sf{\underline{\boxed{\green{\large{\bold{ Solution}}}}}}$

$\sf\implies ( 1 + tan A + sec A ) ( 1 + cot A - cosec A )$

$\sf{\bold{\green{\underline{\underline{tan A = \dfrac{sinA}{cosA}}}}}}$

$\sf\implies ( 1 + \dfrac{sinA}{cosA} + sec A ) ( 1 + cot A - cosec A )$

$\sf{\bold{\green{\underline{\underline{secA = \dfrac{1}{cosA}}}}}}$

$\sf\implies ( 1 + \dfrac{sinA}{CosA}+ \dfrac{1}{cosA} ) ( 1 + cot A - cosec A )$

$\sf{\bold{\green{\underline{\underline{cot A = \dfrac{cosA}{sinA}}}}}}$

$\sf\implies ( 1 + \dfrac{sinA}{CosA}+ \dfrac{1}{cosA} ) ( 1 +\dfrac{cosA}{sinA} - cosec A )$

$\sf{\bold{\green{\underline{\underline{cosecA = \dfrac{1}{sinA}}}}}}$

$\sf\implies(\dfrac{cosA + sin A + 1 }{ cos A})(\dfrac{cos A + Sin A - 1 }{Sin A })$

$\sf{\bold{\green{\underline{\underline{( a+ b ) ( a - b ) = a^2 - b^2 }}}}}$

$\sf\implies\dfrac{ ( cosA + sinA)^2 - 1 }{sinACosA}$

$\sf{\bold{\green{\underline{\underline{(a + b)^2 = a^2 + b^2 + 2ab }}}}}$

$\sf\implies\dfrac{ cos^2A + sin^2A + 2cosAsinA - 1 }{sinACosA}$

$\sf{\bold{\green{\underline{\underline{cos^2A + sin^2A = 1 }}}}}$

$\sf\implies\dfrac{ 1 + 2cosAsinA- 1 }{sinACosA}$

$\sf\implies\dfrac{2cosAsinA}{sinACosA}$

$\sf\longrightarrow 2$

$\sf{\underline{\underline{\large{\bold{\dag (1 + tan A + sec A ) ( 1 + CotA - cosec A ) = 2 }}}}}$