Question

(1+tan theta +sec theta)( 1+cot theta-cosec theta) will be

Answer 1

Solution:

• tan∅ = sin∅/cos∅
• sec∅ = 1/cos∅
• cot∅ = cos∅/sin∅
• csc∅ = 1/sin∅

→ 1 + tan∅ + sec∅ = 1 + sin∅/cos∅ + 1/cos∅

→ 1 + tan∅ + sec∅ = (cos∅ + sin∅ + 1)/cos∅

→ 1 + cot∅ + csc∅ = 1 + cos∅/sin∅ - 1/sin∅

→ 1 + cot∅ + csc∅ = (sin∅ + cos∅ - 1)/sin∅

→ (1 + tan∅ + sec∅)/(1 + cot∅ + csc∅)

→ (cos∅ + sin∅ + 1)/cos∅ × sin∅/(sin∅ + cos∅ - 1)

→ tan∅ × (cos∅ + sin∅ + 1)/(sin∅ + cos∅ - 1)

Multiplying by (cos∅ + sin∅ + 1) we get,

→ tan∅ × (cos∅ + sin∅ + 1)²/[(sin∅ + cos∅)² - 1]

→ tan∅ × (cos²∅ + sin²∅ + 1 + 2cos∅ sin∅ + 2 sin∅ + 2 cos∅)/(sin²∅ + cos²∅ + 2sin∅cos∅ - 1)

→ tan∅ × (1 + 1 + 2cos∅ sin∅ + 2sin∅ + 2cos∅)/(1 + 2sin∅cos∅ - 1)

→ tan∅ × 2(1 + sin∅ + cos∅sin∅ + cos∅)/2(sin∅cos∅)

→ tan∅ × [1/sin∅cos∅ + sin∅/sin∅cos∅ + cos∅/sin∅cos∅ + sin∅cos∅/sin∅cos∅]

→ tan∅(sec∅ csc∅ + sec∅ + csc∅ + 1)

Answer 2

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\huge\boxed{\mathbb{\red{SOLUTION}}}$

To Find The Value of

$\implies(1 + tan \theta + sec \theta)(1 + cot \theta - cosec \theta)=?$

Using ForMuLaS

$\bf\Longrightarrow \tan\theta=\frac{\sin\theta}{\cos\theta}\\</p><p></p><p>\bf\Longrightarrow \cot\theta=\frac{\cos\theta}{\sin\theta}\\</p><p></p><p>\bf\Longrightarrow \cosec\theta=\frac{1}{\sin\theta}\\</p><p></p><p>\bf\Longrightarrow \sec\theta=\frac{1}{\cos\theta}\\</p><p>\bf\Longrightarrow \sin{}^{2}\theta+\cos{}^{2}\theta=1\\</p><p></p><p>\bf\Longrightarrow \sec{}^{2}\theta-\tan{}^{2}\theta=1\\</p><p>\bf\Longrightarrow \cosec{}^{2}\theta-\cot{}^{2}\theta=1\\</p><p>$

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❍$\underline{\red{\mathbb{LET'S \:SOLVE \:THE\: PROBLEM}}}$

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$\implies(1 + tan \theta + sec \theta)(1 + cot \theta - cosec \theta) \\ </p><p></p><p>\implies(1 + \frac{sin \theta}{cos \theta} + \frac{1}{cos \theta} ) (1 + \frac{cos \theta}{sin \theta} - \frac{1}{sin \theta} \\</p><p></p><p> \implies( \frac{cos \theta + sin \theta + 1}{cos \theta} )( \frac{sin \theta \: + cos \theta - 1}{sin \theta}) \\</p><p></p><p> \implies\{ \frac{(sin \theta + cos \theta + 1)(sin \theta + cos \theta - 1)}{sin \theta \: cos \theta} \}\\</p><p></p><p> \implies \frac{ \{(sin \theta + cos \theta) {}^{2} - 1 {}^{2} \}}{sin \theta \: cos \theta} \\ </p><p></p><p> \implies \: \frac{ \{sin {}^{2} \theta + cos {}^{2} \theta + 2sin \theta \: cos \theta - 1 \}}{sin \theta \: cos \theta} \\</p><p></p><p> \implies \: \frac{\cancel1 + 2sin \theta \: cos \theta - \cancel1}{sin \theta \: cos \theta} \: \\ </p><p></p><p>\implies \frac{ \: 2 \: \cancel{sin \theta \: cos \theta}}{\cancel{sin \theta \: cos \theta}} \\ </p><p></p><p> \implies\bf\boxed{\huge{\red{2}}}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

$\mathcal{ \&#35;\mathcal{answer with quality }\: \: \&#38; \: \: \&#35;BAL }$