Question

Proved that Sec A (1 - sin A) (Sec A + tan A) = 1​

Answer 1

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Answer 2

$\underline{ \underline{\large{ \purple{ \bf{to \: prove}}}}}$

$\sf{sec \: A \: (1 - sin \: A)(sec \: A \: + tan \: A) = 1}$

$\underline{ \underline{ \large{ \purple{ \bf{trigonometric \: ratios \: and \: identities \: used}}}}}$

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

$\bullet \sf{tan \theta = \frac{sin \theta}{ cos \theta} }$

$\bullet \sf{1 - {sin}^{2} \theta = {cos}^{2} \theta }$

Algebraic identity used

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

$\underline{ \underline{ \large{ \purple{ \bf{proof}}}}}$

Taking LHS

$\implies \sf{Lhs =sec \: A\: (1 - sin \: A)(sec \: A \: + tan \: A) }$

$\implies \sf{ = \frac{1}{cos \: A}(1 - sin \: A)( \frac{1}{cos \: A} + \frac{sin \: A}{cos \: A} ) }$

$\implies \sf{ = \frac{1 - sin \: A}{cos \: A} \times \frac{1 + sin \: A}{cos \: A} }$

using algebraic identity

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

$\implies \sf{ \frac{ {1}^{2} - {sin}^{2} A }{ {cos}^{2}A} }$

$\implies \sf{ \frac{1 - {sin}^{2}A }{ {cos}^{2} A} }$

using trigonometric identity

$\rm{1 - {sin}^{2} \theta = {cos}^{2} \theta }$

$\implies \sf{ \frac{ {cos}^{2}A }{ {cos}^{2}A } = 1 = Rhs }$

Proved.

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More trigonometric ratios and identites

$\star\rm{cosec\theta=\frac{1}{sin\theta}}$

$\star\rm{cot\theta=\frac{1}{tan\theta}=\frac{cos\theta}{sin\theta}}$

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$\star\rm{1+tan^2\theta=sec^2\theta}$

$\star\rm{1+cot^2\theta=cosec^2\theta}$

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$\star\rm{sin(90-\theta)=cos\theta}$

$\star\rm{cos(90-\theta)=sin\theta}$

$\star\rm{cosec(90-\theta)=sec\theta}$

$\star\rm{sec(90-\theta)=cosec\theta}$

$\star\rm{tan(90-\theta)=cot\theta}$

$\star\rm{cot(90-\theta)=tan\theta}$

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