Question

(sinA + cosA + 1)(sinA-1+ cosA) -secAxcosecA=2 prove​

Answer 1

Given :

${\sf{ (sin A + cos A + 1)(sin A - 1 + cos A) - sec A \times cosecA = 2 }}$

To Prove :

$\text{RHS = LHS}$

Explanation :

${\textsf{We can prove the given equations using below identities}}$

$\sf{\rightarrow(a+b)(a-b)=a^2-b^2}$

$\sf{\rightarrow(a+b)^2 = a^2 + b^2 +2ab}$

$\sf{\rightarrow\cos ^2\theta + \sin ^2\theta = 1}$

$\sf{\rightarrow\sin \theta = \frac{1 }{\csc \theta}}$

$\sf{\rightarrow\cos \theta = \frac{1}{\sec \theta}}$

Solution :

${\sf{= (sinA + cosA +1)(sinA + cosA -1)- (secA \times cosec A)}}$

${\sf{= ((sin A + cos A)+1)((sin A+ cosA )-1)- (secA \times CosecA)}}$

${\sf{= ((sinA + cosA)^2 -1^2) (secA \times cosecA)}}$

${\sf{(\because((a+b)(a-b)=a^2 - b^2 )) \:in \:which\:a = sinAcosA \:\&\: b = 1 ) }}$

${\sf{= ((sin^A + cos^2 A +2cosAsinA)-1) ( secA \times cosecA))}}$

${\sf{(\because (a+b)^2 = (a^2+b^2+2ab)\: in\:which\:a=sinA\:\&\:b=cosA)}}$

$= {\sf{(1+2cosAsinA -1 )(secA\times cosecA)}}$

$= {\sf{2sinAcosA \times secA \times cosecA}}$

$= {\sf{2 \times cosA \times sinA \times \frac{1}{sinA}\times \frac{1}{cosA}}}$

${\sf{(\because secA = \frac{1}{cosA} \: \&\:cosecA = \frac{1}{sinA})}}$

$= {\sf{\frac{2 \times sinA \times cosA}{sinA \times cosA}}} \implies 2$

$\sf{Hence\:\: proved}$

Answer 2

${\large{\underbrace{\underline{\bf{Question:-}}}}}$

Q)) Prove that :

$\rm \: (sin \: A + cos \: A + 1)(sin \: a - 1 + cos \: A) \times sec \: A \times cosec \: A = 2$

${\large{\underbrace{\underline{\bf{Answer\;\checkmark}}}}}$

$\bf \: LHS : \\ \\ \rightarrow \sf \: (sin \: A + cos \: A + 1)(sin \: A - 1 + cos \: A) \times sec \: A \times cosec \: A \\ \\ \rightarrow \sf \: \{( sin \: A + cos \: A) + 1 \} \{(sin \: A+ cos \: A) - 1 \} \times sec \: A \times cosec \: A \\ \\ \rightarrow \sf \: \{ {(sin \: A + cos \: A)}^{2} - {1}^{2} \}\times sec \: A \times cosec \: A \\ \\ \rightarrow \sf \: \{ \blue{\underline{{sin}^{2} \: A + {cos}^{2} \: A}} + 2sin \: A \times cos \: A - 1 \} \times sec \: A \times cosec \: A\\ \\ \rightarrow \sf \: \{ \cancel1 + 2sin \: A\times cos \: A - \cancel1 \} \times sec \: A \times cosec \: A \\ \\ \rightarrow \sf \: 2sin \: A \times cos \: A \times sec \: A \times cosec \: A\\ \\ \rightarrow\sf\:2\times \cancel{sin\:A}\times\cancel{cos\:A}\times\frac{1}{\cancel{cos\:A}}\times\frac{1}{\cancel{sin\:A}}\\ \\ \sf\rightarrow 2\times1\times1 \\ \\ \colon\implies\:\underline{\boxed{\tt\pink{LHS=2=RHS}}}\:\rm\purple{Hence\:Proved}$

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${\underline{\underline{\rm{Formula\:Used:}}}}$

${ \boxed{\begin{array}{ c }{ \sf \star \: (x + y)(x - y) = {x}^{2} - {y}^{2} }\\ \\ { \sf \star \: {sin}^{2} \: \alpha + {cos}^{2} \: \alpha = 1 } \\ \\ { \star \sf \: cosec \: \alpha = \dfrac{1}{sin \: \alpha } } \\ \\ { \star \sf \: sec \: \alpha = \dfrac{1}{cos \: \alpha } }\end{array}}}$

${\underline{\underline{\rm{Know\:More:}}}}$

$\boxed{ \begin{array}{c}{ \sf \: \bull \: {sin}^{2} \: \alpha + {cos}^{2} \: \alpha = 1} \\ \\ \bull \sf \: \: {tan}^{2} \: \alpha + 1 = {sec}^{2} \: \alpha \\ \\ \bull \sf \: \: 1 + {cot}^{2} \: \alpha = {cosec}^{2} \: \alpha \\ \\ \bull \sf \: \: cosec \: \alpha = \dfrac{1}{sin \: \alpha } \\ \\ \bull \sf \: \: sec \: \alpha = \dfrac{1}{cos \: \alpha } \\ \\ \sf \bull \: \: cot \: \alpha = \dfrac{1}{tan \: \alpha } \end{array}}$