Question

Show that, tan^theta + cot^theta + 2 = sec^thata ×cosec^theta
plz answer fast​

Answer 1

$\bigstar\large\underline{\rm{\sf Correct \:Question}}$

Show that:-

$\sf\tan\theta+\cot\theta+2=\sec\theta.\cosec\theta\red{+2}$

$\bigstar\large\underline{\rm{\sf Required\:Solution}}$

$:\implies\sf \tan\theta+\cot\theta+2$

$:\implies\sf \dfrac{\sin\theta}{\cos\theta} +\dfrac{\cos\theta}{\sin\theta}+2$

$:\implies\sf \dfrac{\sin^{2} \theta+\cos^{2} \theta}{\sin\theta.\cos\theta}+2$

As we know that:-

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

$:\implies\sf \dfrac{1}{\sin\theta.\cos\theta} +2$

$:\implies\sf\red{\sec\theta.\cosec\theta+2}$

ADDITIONAL KNOWLEDGE:-

Important Identities:-

• sec²∅ - tan²∅ = 1
• cosec²∅ - cot²∅ = 1

Reciprocal of ratios:-

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

Ratios of Complemantary Angles:-

• sin∅ = cos(90 - ∅)
• cos∅ = sin(90 - ∅)
• tan∅ = cot(90 - ∅)
• cot∅ = tan(90 - ∅)
• sec∅ = cosec(90 - ∅)
• cosec∅ = sec(90 = ∅)

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I hope it's beneficial :D