Question

Prove that ( 1 + cotA- cosecA) ( 1 + tanA + secA) =2​

Answer 1

$\huge{\underline{ \red{\frak{Solution : - }}}}$

$:\implies \sf \bigg(1 + \cot(A) - \cosec(A) \bigg) \bigg(1 + \tan(A) + \sec(A) \bigg) = 2$

L.H.S :-

$:\implies \sf \Bigg \lgroup1 + \dfrac{\cos(A) }{\sin(A)}- \dfrac{1}{\sin(A)} \Bigg \rgroup \Bigg \lgroup1 + \dfrac{\sin(A) }{\cos(A)}- \dfrac{1}{\cos(A)} \Bigg \rgroup$

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

$:\implies \sf \dfrac{ \bigg(\sin(A) + \cos(A)\bigg)^{2} - 1}{\sin(A).\cos(A)} \qquad \qquad \Bigg\lgroup \textsf{\textbf{ \because a ^ \text2 - b ^ \text2 = (a + b) (a - b)}}\Bigg\rgroup$

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

$:\implies \sf \dfrac{1 - 1 + 2\sin(A)\cos(A)}{\sin(A).\cos(A)}\qquad \qquad \Bigg\lgroup \textsf{\textbf{ \because \sin ^ \text2 - \cos ^ \text2 = 1}}\Bigg\rgroup$

$:\implies \sf \dfrac{ 2\sin(A)\cos(A)}{\sin(A).\cos(A)}$

$:\implies \underline{ \boxed{ \sf 2= RHS }}$

ஃ LHS = RHS

Hence Proved

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$\boxed{\boxed{\begin{minipage}{6cm} \textsf{\textbf{ Important Trigonometric identities :-}} \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}}$