Question

Prove that: (1+cotA-cosecA)(1+tanA+secA)=2​

Answer 1

Taking LHS,

$\longrightarrow LHS=(1+\cot A-\csc A)(1+\tan A+\sec A)$

We can take,

• $\cot A=\dfrac{\cos A}{\sin A}$

• $\csc A=\dfrac{1}{\sin A}$

• $\tan A=\dfrac{\sin A}{\cos A}$

• $\sec A=\dfrac{1}{\cos A}$

Then,

$\longrightarrow LHS=\left(1+\dfrac{\cos A}{\sin A}-\dfrac{1}{\sin A}\right)\left(1+\dfrac{\sin A}{\cos A}+\dfrac{1}{\cos A}\right)$

$\longrightarrow LHS=\dfrac{(\sin A+\cos A-1)(\cos A+\sin A+1)}{\sin A\cos A}$

$\longrightarrow LHS=\dfrac{((\sin A+\cos A)-1)((\sin A+\cos A)+1)}{\sin A\cos A}$

Since $(a-b)(a+b)=a^2-b^2,$

$\longrightarrow LHS=\dfrac{(\sin A+\cos A)^2-1^2}{\sin A\cos A}$

$\longrightarrow LHS=\dfrac{\sin^2A+\cos^2A+2\sin A\cos A-1}{\sin A\cos A}$

Since $\sin^2A+\cos^2A=1,$

$\longrightarrow LHS=\dfrac{1+2\sin A\cos A-1}{\sin A\cos A}$

$\longrightarrow LHS=\dfrac{2\sin A\cos A}{\sin A\cos A}$

$\longrightarrow LHS=2$

$\longrightarrow LHS=RHS$

Hence Proved!

Answer 2

See the photo attached for solution