Question

(cosec theta - sin theta) ( sec theta - cos theta) (tan theta +cot theta) is equal to

a)0
b)1
c)-1
d)none of these ​

Answer 1

Answer:

Option b) 1

Step-by-step explanation:

$(\cosec\theta - \sin\theta) (\sec\theta - \cos\theta)(\tan\theta + \cot\theta)$

We know that,

$\: \: \: \: \: \: \cosec = \dfrac{1}{\sin}$

$\longrightarrow \:( \frac{1}{\sin} \theta - \sin\theta) (\sec\theta - \cos\theta)(\tan\theta + \cot\theta)$

We know that,

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

$\longrightarrow \:( \frac{1}{\sin} \theta - \sin\theta) ( \frac{1}{\cos} \theta - \cos\theta)(\tan\theta + \cot\theta)$

We know that,

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

$\: \: \: \: \: \: \cot = \dfrac{\cos}{\sin}$

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

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

We know that,

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

$\longrightarrow \:\frac{ {\cos}^{2} \theta}{\sin\theta} \times \frac{ {\sin}^{2}\theta}{\cos\theta} \times \frac{1}{\cos\theta \: \sin\theta}$

$\longrightarrow \: {\cos} \theta\times {\sin}\theta \times \frac{1}{\cos\theta \: \sin\theta}$

$\longrightarrow \: 1$

Therefore, answer is 1.

Answer 2

Answer :-

(b) 1

Explanation :-

• $\sf (cosec ∅ - sin ∅ ) (sec ∅ - cos∅ ) (tan ∅ + cot∅ )$

• As we know that,

• $\sf cosec ∅ = \frac{1}{sin∅}$

• $\sf sec∅ = \frac{1}{cos∅}$

• $\sf tan ∅ = \frac{sin∅}{cos∅}$

• $\sf cot ∅ = \frac{cos∅}{sin∅}$

Now, put the values

• $\sf (cosec ∅ - sin ∅ ) (sec ∅ - cos∅ ) (tan ∅ + cot∅ )$

• $\sf \bigg(\frac{1}{sin∅} - sin∅ \bigg) \bigg(\frac{1}{cos∅} - cos∅ \bigg) \bigg(\frac{sin∅}{cos∅}+ \frac{cos∅}{sin∅} \bigg)$

• $\sf \bigg( \frac{1 - {sin}^{2} ∅ }{sin∅ } \bigg) \bigg( \frac{ 1 - {cos}^{2}∅ }{cos∅ } \bigg) \bigg( \frac{ {sin}^{2} ∅ + {cos}^{2} ∅ }{sin∅ \times cos∅ } \bigg)$

• also we know that, sin² ∅ + cos²∅ = 1

• sin²∅ = 1 - cos²∅

• cos²∅ = 1 - sin² ∅

Hence,

• $\sf \bigg( \frac{ {cos}^{2}∅ }{sin∅} \bigg) \bigg( \frac{ {sin}^{2} ∅}{cos∅} \bigg) \bigg( \frac{ {sin}^{2}∅ + {cos}^{2} ∅}{sin∅ \times cos∅} \bigg)$

• $\sf \bigg( \frac{ {cos}^{ \cancel2}∅ }{ \cancel{sin∅}} \bigg) \bigg( \frac{ {sin}^{ \cancel2} ∅}{ \cancel{cos∅}} \bigg) \bigg( \frac{ {sin}^{2}∅ + {cos}^{2} ∅}{sin∅ \times cos∅} \bigg)$

• $\sf \cancel{cos∅ \times sin∅} \bigg( \frac{ {sin}^{2} ∅ + {cos}^{2}∅ }{ \cancel{cos∅ \: \: sin∅} } \bigg)$

• $\sf sin² ∅+ cos² ∅$

• $\sf = \fbox{ 1 }$

For more information :-

• Sin² ∅ + Cos² = 1
• Sin² ∅ = 1 - Cos² ∅
• Cos² ∅ = 1 - Sin² ∅

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• Sec² ∅ - Tan² ∅ = 1
• Sec² ∅ = 1 + Tan² ∅
• Tan² ∅ = 1 + Sec² ∅

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• Cosec² ∅ - Cot² ∅ = 1
• Cosec² ∅ = 1 + Cot² ∅
• Cot² ∅ = 1 + Cosec² ∅

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I hope it helps you ❤️✔️