Question

prove that: sin(90^-A).cos(90^-A)= cot A/1+cot Square A

Answer 1

Step-by-step explanation:

Given :-

Sin (90°-A) Cos (90°-A)

To find :-

Prove that

Sin (90°-A) Cos (90°-A)=CotA/(1+Cot²A)

Solution :-

On taking LHS :

Sin (90°-A) Cos (90°-A)

We know that

Sin (90°-A) = Cos A

Cos (90°-A) = Sin A

Sin (90°-A) Cos (90°-A) = Cos A Sin A ------------(1)

On taking RHS

CotA/(1+Cot²A)

We know that

Cosec² A - Cot² A = 1

=> Cosec² A = 1 + Cot² A

Now

CotA/(1+Cot²A)

=> Cot A / Cosec² A

=> CotA × Sin² A

Since , Cosec A = 1/Sin A

=> (Cos A / Sin A ) × Sin² A

Since , CotA = Cos A / Sin A

=> (Cos A × Sin² A )/Sin A

=> Cos A . Sin A ----------------(2)

From (1) &(2)

LHS = RHS

Sin (90°-A) Cos (90°-A)=CotA/(1+Cot²A)

Hence, Proved.

Used formulae:-

→ Sin (90°-A) = Cos A

→ Cos (90°-A) = Sin A

→ Cosec² A - Cot² A = 1

→ Cosec A = 1/Sin A

→ CotA = Cos A / Sin A