Question

Prove the following relations.
Cos²A + cos²A.cot²A=cot²A
(1-sin²A)(1+cot²A)=cot²A​

Answer 1

Step-by-step explanation:

Let's solve for the first one,

$R.H.S. = Cos²A + cos²A.cot²A$

Step 1- Take Cos²A common,

$= Cos²A( 1+ cot²A)$

$\small \sf \: Now \: we \: know \: that \\ \small (1 + Cot^{2} \theta = Cosec^{2} \theta)$

Step 2 - Substituting above value in step 1,

$= Cos^{2} A(Cosec^2A)$

$\small \sf \: We \: know \: that \: Cosec \theta= \frac{1}{Sin \theta}$

Step 3 - Substitute Cosec²A with 1/Sin²A

$= \frac{Cos^2A}{Sin^2A}$

$\small \sf \: Now \: the \: basic \: trigonometric \: function \: is \\ \small \frac{Cos\theta}{Sin \theta} = Cot\theta$

Step 4 - Replacing the product of step 3 with above function,

$= {Cot}^{2} A$

Hence proved,

$R.H.S = L.H.S$

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Now let's prove the second relation,

$(1-sin²A)(1+cot²A)=cot²A$

Let's solve for R.H.S.,

$R.H.S.= (1-sin^{2} A)(1+cot^{2} A)$

As I told above,

$\small(1+cot^{2} \theta) = cosec^{2} \theta$

Another important relation is,

$\small sin^2\theta + cos^2\theta= 1 \: or \: \\ \small cos^2\theta = 1 - sin^2\theta$

Replacing both the above brackets with this relation we get,

$= Cos^2A \cdot Cosec^2A$

Now follow the step 3 and step 4 of first question and you will get,

$R.H.S. = L.H.S.$

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Some important trigonometric relations that you must learn,

$Sin^2\theta + Cos^2\theta = 1 \\ 1 + Cot^2\theta = Cosec^2\theta \\ 1+Tan^2\theta = Sec^2\theta$

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