Question

Prove that (1+cot teta)² +(1- cot teta)² = 2 cosec²teta​

Answer 1

Step-by-step explanation:

Answer:-

$(1 + \sf{cot \: \theta})^{2} + (1 - \sf{cot \: \theta})^{2}$

So, breaking the Brackets under:-

$\sf{(x + y)^{2} = {x}^{2} + {y}^{2} + 2xy}$

$\sf{(x - y)^{2} = {x}^{2} + {y}^{2} - 2xy}$

Let's Do!

$\sf{ cot^{2} \: \theta + 1 + 2 \: cot \: \theta} + \sf{ cot^{2} \: \theta + 1 - 2 cot \: \theta}$

Extend this side $\longrightarrow$

$\sf{ cot^{2} \: \theta + 1 + \cancel{ 2 \: cot \: \theta}} + \sf{ cot^{2} \: \theta + 1 \cancel{ - 2 cot \: \theta}}$

$= \sf{ 2cot^{2} \: \theta +2 }$

$= \sf{ 2(cot^{2} \: \theta + 1) }$

Now,

$= \sf{ cot^{2} \: \theta + 1 = cosec ^{2} \theta }$

So,

$= \sf{ 2 \: cosec^{2} \: \theta }$

So, LHS = RHS

Hence Proved!

Answer 2

Answer :-

★ Concept :-

Here the concept of Trignometric Ratios has been used. According to this, the value of opposite pair of Trignometric identities depend on each other.

Now first we will simplify LHS and then we will go for RHS which will give final equal values as output.

★ Solution :-

• To Prove :-

➫ (1 + cot θ)² + (1 - cot θ)² = 2 cosec² θ

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• Identities to be used :-

➥ (x + y)² = x² + y² + 2xy

➥ (x - y)² = x² + y² - 2xy

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• Proof :-

Let us take LHS first,

➠ 1 + cot² θ + 2cotθ + 1 + cot² θ - 2cotθ

Cancelling the like terms we get,

➠ 2 + 2cot² θ

Taking two in common we get,

➠ 2(1 + cot² θ)

By identity we know that,

✒ cosec² θ = 1 + cot² θ

By applying this, we get,

➠ 2 cosec² θ = LHS

Clearly, LHS = RHS = 2 cosec² θ

Hence, proved ..

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★ More to know :-

• Trignometric relations are given as ,

✏ sin² θ + cos² θ = 1 [vicé - versa]

✏ cosec² θ = 1 + cot² θ [vicé - versa]

✏ sec² θ = 1 + tan² θ [vicé - versa]

• Trignometric relations are given as,

✏ sin θ = Perpendicular / Hypotenuse

✏ cos θ = Base / Hypotenuse

✏ tan θ = Perpendicular / Base

✏ cosec θ = 1 / (sin θ)

✏ sec θ = 1 / (cos θ)

✏ cot θ = 1 / (tan θ)