Question

Prove this equation ​

Answer 1

To prove:

$\dfrac{sin\theta\; tan\theta}{1 - cos\theta} = 1 + sec\theta$

Proof:

RHS = $1 + sec\theta$

LSH = $\dfrac{sin\theta\; tan\theta}{1 - cos\theta}$

Rationalizing the denominator:

LHS = $\dfrac{sin\theta\; tan\theta}{1 - cos\theta} \times \dfrac{1+cos\theta}{1+cos\theta}$

$= \dfrac {sin\theta\; tan\theta(1+cos\theta)}{{sin}^{2}\theta}$

$=\dfrac {tan\theta (1+cos\theta)}{sin\theta}$

$=\dfrac {sin\theta(1+cos\theta)}{cos\theta\;sin\theta}$

$=\dfrac{1+cos\theta}{cos\theta}$

RHS = $1 + sec\theta$

$=1+ \dfrac{1}{cos\theta}$

$=\dfrac{cos\theta + 1}{cos\theta}$

RHS = LHS

Hence proved.

Identities used:

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

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

$tan\theta = \dfrac{sin\theta}{cos\theta}$

$sec\theta = \dfrac{1}{cos\theta}$

Answer 2

Solution:-

To Proof:-

[ ( sinA . tanA )/ ( 1- cosA) ] =( 1 + secA )

Proof :-

L.H.S.

[ ( sinA . tanA )/ ( 1- cosA) ]

On Rationalizing. we get,

=) [ ( sinA . tanA )/ ( 1- cosA) ] × [ ( 1 +cosA)/ ( 1 + cosA )]

=) [ ( sinA . tanA ). ( 1 + cosA)/ ( 1- cosA).( 1 + cosA) ]

=)[ ( sinA . tanA ). ( 1 + cosA)/ ( 1² - cosA² ) ]

=) [ ( sinA . tanA ). ( 1 + cosA)/ sin²A ]

=) [ tanA. ( 1 +cosA)/ sinA ]

=) [ sinA/cosA . ( 1 + cosA)/sinA ]

=) [ ( 1 + cosA)/cosA ]

=) [ 1/cosA + cosA/cosA ]

=) [ secA + 1 ]

=) 1 + secA.

Hence Proved!

Identity Used:-

• ( a² - b² ) = ( a + b) ( a - b )

Trigonometric Formulas:-

• ( sin²A + cos²A ) = 1
• ( sec²A - tan²A ) = 1
• ( cosec²A - cot²A ) = 1