Question

1+sec/sec=sin^2/1-cos​

Answer 1

Answer:

${ \underline{ \large{ \pmb{ \sf{Question : }}}}}$

• ${ \sf{Prove \: that : \frac{1 + sec}{sec} = \frac{ {sin}^{2} }{1 - cos} }}$

${ \underline{ \large{ \pmb{ \sf{Solution:}}}}}$

$: { \implies{ \sf{ \frac{1 + secA}{secA} }}} \\ \\ : { \implies{ \sf{ \frac{1 + \frac{1}{cosA} }{ \frac{1}{cosA} } }}} \\ \\ : { \implies{ \sf{ 1 + \frac{1 }{cosA} \times \frac{cosA}{1} }}} \\ \\ : { \implies{ \sf{ \frac{cosA + 1}{cosA} \times cosA}}} \\ \\ : { \implies{ \sf{ \frac{cosA + 1}{{ \cancel{cosA}}} \times { \cancel{cosA}}}}} \\ \\ : { \implies{ \sf{cosA + 1}}} \\ \\ { \pmb{ \sf{Multiply \: And \: Divide \: With \: (1- CosA)}}} \\ \\ : { \implies{ \sf{cosA + 1 \times \frac{1 - cosA}{1 - cosA} }}} \\ \\ : { \implies{ \sf{ \frac{1 - {cos}^{2}A }{1 - cosA} }}} \\ \\ : { \implies{ \sf{ \frac{ {sin}^{2}A }{1 - cosA} }}} \\ \\ \therefore { \pmb{ \sf{Hence \: Proved}}}$

Used Formulae:

• (a² - b²) = (a + b) (a - b)

Used Identities:

• Sin² + cos² = 1
• Sin² = 1 - cos²

Answer 2

$\bullet\large{\sf{\underline{\underline\red{Solution!!}}}}$

• L.H.S = R.H.S

Step-by-step explanation:

According to this question

Given that,

$\tt\frac{1 + sec a}{sec a}$

$\tt\frac{sin 2a}{1 - cos a}$

Let, L.H.S

$\tt\frac{1 + sec a}{sec a}$

$\tt\frac{1 + (\frac{1}{cos a})}{\frac{1}{cos a}}]$

$\large\tt\frac{\frac{cos a + 1}{cos a}}{\frac{1}{cosa}}$

cos a + 1cosa+1

Multiple and divide with (1 - cos a)

$\tt(1 + cos a)\times \frac{1 - cos a}{1 - cos a}$

$\tt\frac{1 - cos^2 a}{1 - cos aa}$

$\tt\frac{sin 2a}{1 - cos a}$

R.H.S