Question

$\large\sf Prove \: that: \\ \\ \ \textless \ br /\ \textgreater \ \tt (1+ { \cot}^{2} )(1+ \cos A)(1- \cos A)=1$
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Answer 1

EXPLANATION.

$\sf \: equation \: are \\ \\ \sf : \implies \: (1 + { \cot }^{2}A)(1 + \: \cos \: A) (1 - \cos \: A) = 1$

$\sf : \implies \: 1 + \cot {}^{2} (A) = \csc {}^{2} (A) \\ \\ \sf : \implies \: 1 - \cos {}^{2} (A) = \sin {}^{2} (A) \\ \\ \sf : \implies \: using \: identity \: = (x + y)(x - y) = ( {x}^{2} - {y}^{2} )$

$\sf : \implies \: (1 + \cot {}^{2} A)(1 - \cos \:A )(1 + \cos \: A) \\ \\ \sf : \implies \: ( \csc {}^{2} A)(1 - \cos {}^{2} A) \\ \\ \sf : \implies \: \frac{1}{ ( \cancel{\sin {}^{2} \:A ) }} \times \cancel{\sin \: {}^{2} A} = 1$

$\sf : \implies \: \green{{ \underline{some \: related \: formula}}} \\ \\ \sf : \implies \: \sin {}^{2} ( \theta) + \cos {}^{2} ( \theta) = 1 \\ \\ \sf : \implies \: 1 + \tan {}^{2} ( \theta) = \sec {}^{2} ( \theta) \\ \\ \sf \implies \: 1 + \cot {}^{2} ( \theta) = \csc {}^{2} ( \theta)$

$\sf : \implies \: 1 - \cos(2 \theta) = 2 \sin {}^{2} ( \theta) \\ \\ \sf : \implies \: 1 + \cos( 2\theta) = 2 \cos {}^{2} ( \theta) \\ \\ \sf : \implies \: 1 - \sin(2 \theta) = ( \cos\theta \: - \sin\theta) {}^{2} \\ \\ \sf : \implies \: 1 + \sin(2 \theta) = ( \cos \theta + \sin \theta) {}^{2}$

Answer 2

$\bf{\underline{\large{\: To \: Prove }}}\\ \\ \bf{ \bigstar \: (1 + { \cot }^{2}A)(1 + \: \cos \: A) (1 - \cos \: A) = 1}$

Taking L.H.S

$\rm \longrightarrow \: (1 + \cot {}^{2} A)(1 - \cos \:A )(1 + \cos \: A)$

$\underline{ \boxed{ \sf{Using \: identity \: = (x + y)(x - y) = ( {x}^{2} - {y}^{2} )}}} \\ \\ \rm \longrightarrow \: ( 1 + \cot {}^{2} A)(1 - \cos {}^{2} A)$

$\underline{\frak{We \: know \: that \: }}\begin{cases} \sf \bullet \: 1 + cot {}^{2} (A) = csc {}^{2} (A) \\ \sf \bullet \: 1 - cos {}^{2} (A) = sin {}^{2} (A) \\ \sf \: \bullet \: csc (A) = \dfrac{1}{sin(A)} \end{cases}$

$\rm{ \longrightarrow \: csc {}^{2} A \times \sin \: {}^{2} A } \\ \\ \rm \longrightarrow \: \dfrac{1}{ ( \cancel{\sin {}^{2} \:A ) }} \times \cancel{\sin \: {}^{2} A} \\ \\ \rm \longrightarrow \: 1$

L.H.S = R.H.S ( HENCE PROVED )