Question

sin a-cos a + 1 upon sin a +cos a -1 is equal to 1 upon sec a -tana

Answer 1

Solution

→ (sinA - cosA + 1)/(sinA + cosA - 1)

→ (sinA - cosA + 1)/(sinA + cosA - 1) × (sinA + cosA + 1)/(sinA + cosA + 1)

→ (sina - cosA + 1)(sinA + cosA + 1)/[(sinA + cosA)² - 1²]

→ (sin²A + sinA cosA + sinA - sinA cosA - cos²A - cosA + sinA + cosA + 1)/(sin²A + cos²A + 2 sinA cos - 1)

→ (sin²A + 2 sinA - cos²A + 1)/(1 + 2 sinA cosA - 1)

→ (sin²A + 2 sinA - 1 + sin²A + 1)/(2 sinA cosA)

→ (2 sin²A + 2 sinA)/2sinA cosA

→ 2sinA(sinA + 1)/2 sinA cosA

→ (sinA + 1)/cosA

→sinA/cosA + 1/cosA

→ tanA + secA

Multiplying by (secA - tanA)/(secA - tanA)

→ (tanA + secA)/1 × (secA - tanA)/(secA - tanA)

→ (sec²A - tan²A)/(secA - tanA)

→ 1/(secA - tanA)

Hence Proved

Answer 2

$\LARGE\underline{\underline{\sf \red{T}\blue{o}\:\green{P}\orange{r}\pink{o}\red{v}\blue{e}:}}$

(sin a-cos a + 1)/(sin a +cos a -1) =1/(sec a -tana)

$\LARGE\underline{\underline{\sf \red{S}\blue{o}\green{l}\orange{u}\pink{t}\purple{i}\orange{o}\red{n}:}}$

$\textsf{Formula Used:-}$

• sec²a - tan²a = 1
• (a²-b²) = (a+b)(a-b)
• sina/cosa = tana
• 1/cosa = seca

$\textsf{See my solution in easiest way now :--}$

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$\red{\bold{\underline{\underline{LHS=}}}}$

$\frac{ \sin(a) - \cos(a) + 1}{\sin(a) + \cos(a) - 1} \\ \\ dividing \: both \: numerator \: and \: denominator \: \\ by \: cosa \: we \: get \: \\ \\ \frac{ \tan(a) - 1 + \sec(a) }{\tan(a) + 1 - \sec(a)} \\ \\ putting \: value \: of \: 1 \: in \: denominator \\ as \: \sec ^{2}a - \tan^{2}a \: we \: get \\ \\ \frac{\tan(a) - 1 + \sec(a) }{(\tan(a) - \sec(a) + \sec ^{2}a - \tan^{2}a)} \\ \\ \frac{ \cancel{\tan(a) - 1 + \sec(a)}}{( \sec(a) - \tan(a) )( \cancel{- 1 + \sec(a) + \tan(a) )} } \\ \\ \frac{1}{\sec(a) - \tan(a)}$

$\red{\bold{\underline{\underline{=RHS}}}}$

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$\huge\underline\mathfrak\green{Hence\:Proved}$

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$\color {red}\large\bold\star\underline\mathcal{Extra\:Brainly\:Knowledge:-}$

$\underline\textsf{Double Angle Identities}$

→ sin²θ = 2sinθcosθ

= $\frac{2 \tanθ}{1+ \tan^{2}θ}$

→ cos2θ = cos²θ- sin²θ

= 1 - 2sin²θ

= 2cos²θ- 1

= $\frac{1- \tan^{2}θ}{1+ \tan^{2}θ}$

→Tan2θ = $\frac{2 \tanθ}{1- \tan^{2}θ}$

$\mathcal{BE\:\:BRAINLY}$