If $$(sin⁡β+cosec⁡β)^2+(sec⁡β-cos⁡β)^2=a+tan^2⁡β+cot^2⁡β,$$​ then the value of a is:

Correct option is A

Given:

​$$(\sin \beta + \cosec \beta)^2 + (\sec \beta - \cos \beta)^2 = a + \tan^2 \beta + \cot^2 \beta$$ 

Value of a = ?

Concept Used:

​$$sinβ⋅cosecβ=1$$ 

​$$secβ⋅cosβ=1$$

$$\sin^2 \beta + \cos^2 \beta = 1$$ 

$$\cosec^2 \beta = 1 + \cot^2 \beta$$ 

​​$$\sec^2 \beta = 1 + \tan^2 \beta$$​​

Solution: 

​$$(\sin \beta + \cosec \beta)^2 + (\sec \beta - \cos \beta)^2 = a + \tan^2 \beta + \cot^2 \beta$$ 

By expanding LHS

​$$\sin^2 \beta + 2 \sin \beta \cosec \beta + \cosec^2 \beta + \sec^2 \beta - 2 \sec \beta \cos \beta + \cos^2 \beta = a + \tan^2 \beta + \cot^2 \beta$$ 

$$\sin^2 \beta + \cos^2 \beta + 2\sin\beta\cdot\cosec\beta -2\sec\beta\cdot\cos\beta + \cosec^2 \beta + \sec^2 \beta = a + \tan^2 \beta + \cot^2 \beta$$ 

​$$1+2-2 + \cosec^2 \beta + \sec^2 \beta = a + \tan^2 \beta + \cot^2 \beta$$ 

​$$1+ \cosec^2 \beta + \sec^2 \beta = a + \tan^2 \beta + \cot^2 \beta$$ 

Using identity of $$\cosec^2\beta\quad\text{and}\quad\sec^2\beta$$ 

​$$1+ 1+ \cot^2 \beta + 1+ \tan^2 \beta = a + \tan^2 \beta + \cot^2 \beta$$​​

​$$3+\cot^2 \beta+\tan^2 \beta = a + \tan^2 \beta + \cot^2 \beta$$ 

By compering both sides; 

a = 3 

Thus, the value of $a$ is 3.​