Question

(1+tano+seco)(1+coto-coseco= (a) 0 (b) 1 (c) 2 (d) -1​

Answer 1

Answer:

-1..........AJUABHIA 94

Answer 2

$\sf\huge\blue{\underline{\underline{ Question : }}}$

(1 + tan θ + sec θ)(1 + cot θ + cosec θ) = ?

A) 0

B) 1

C) 2

D) - 1

$\sf\huge\blue{\underline{\underline{ Solution : }}}$

$\bf\:\implies (1 + tan\: \theta + sec\: \theta)(1 + cot \:\theta - cosec\:\theta)$

• tan θ = sin θ/cos θ
• sec θ= 1/cos θ
• cot θ = cos θ/sin θ
• cosec θ = 1/sin θ

$\bf\:\implies (1 + \frac{ \sin\:\theta}{\cos\:\theta}+ \frac{1}{\cos\:\theta})(1 + \frac{\cos\:\theta}{\sin\:\theta} - \frac{1}{\sin\:\theta})$

$\bf\:\implies (1 + \frac{ \sin\:\theta}{\cos\:\theta}+ \frac{1}{\cos\:\theta})(1 + \frac{\cos\:\theta}{\sin\:\theta} - \frac{1}{\sin\:\theta})$

$\bf\:\implies (\frac{ (\sin\:\theta + \sin\:\theta) + (1)}{\cos \:\theta})(\frac{ (\sin\:\theta + \cos\:\theta ) -1}{\sin\:\theta})$

• [ (a + b)(a - b) = a² - b² ]

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

• [ (a + b)² = a² + b² + 2ab ]

$\bf\:\implies \frac{\sin^{2}\:\theta + \cos^{2}\:\theta + 2\sin\:\theta\cos\:\theta - 1 }{\sin\:\theta\cos\:\theta}$

• [ sin² θ + cos² θ = 1 ]

$\bf\:\implies \frac{1 + 2\sin\:\theta\cos\:\theta - 1}{\sin\:\theta\cos\:\theta}$

$\bf\:\implies \frac{2\sin\:\theta\cos\:\theta}{\sin\:\theta\cos\:\theta}$

$\bf\:\implies 2$

$\underline{\boxed{\bf{\purple{ \therefore (1 + \tan\:\theta + \sec\:\theta)(1 + \cot\:\theta- \cosec\:\theta) = 2 (C) }}}}\:\orange{\bigstar}$

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★ Trigonometric Identities :

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

$\rm\red{\implies \sec^{2} \theta - \tan^{2} \theta = 1 }$

$\rm\red{\implies \csc^{2} \theta - \cot^{2} \theta = 1 }$

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