Question

If a = secx-tanx and b = cosecx+cotx, then show that ab+a-b+1 =0.

Answer 1

Hint: Convert the expression $ab+a-b+1$ in terms of x. Hence convert the resulting expression in terms of sines and cosines. Take sinxcosx as L.C.M in the resulting expression and expand the products in the numerator. Use the fact that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to simplify the expression and hence prove that ab+a-b+1 =0.

Complete step-by-step answer:
Let S = ab+a-b+1
Substituting the values of a and b , we get
S = (secx-tanx)(cosecx+cotx)+secx-tanx-(cosecx+cotx) +1
Converting to sines and cosines using $\sec x=\dfrac{1}{\cos x},\csc x=\dfrac{1}{\sin x},\cot x=\dfrac{\cos x}{\sin x}$ and $\tan x=\dfrac{\sin x}{\cos x}$, we get
S $=\left( \dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x} \right)\left( \dfrac{1}{\sin x}+\dfrac{\cos x}{\sin x} \right)+\left( \dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x} \right)-\left( \dfrac{1}{\sin x}+\dfrac{\cos x}{\sin x} \right)+1$
Hence, we have
S $=\dfrac{1-\sin x}{\cos x}\times \dfrac{1+\cos x}{\sin x}+\dfrac{1-\sin x}{\cos x}-\dfrac{1+\cos x}{\sin x}+1$
Taking sinxcosx as L.C.M., we get
S $=\dfrac{\left( 1-\sin x \right)\left( 1+\cos x \right)+\left( 1-\sin x \right)\sin x-\left( 1+\cos x \right)\cos x}{\sin x\cos x}+1$
Expanding using the distributive property of multiplication over addition and subtraction, we get
S $=\dfrac{1-\sin x+\cos x-\cos x\sin x+\sin x-{{\sin }^{2}}x-\cos x-{{\cos }^{2}}x}{\sin x\cos x}+1$
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
Using the above identity, we get
S $=\dfrac{1-1-\cos x\sin x}{\cos x\sin x}+1$
Hence, we have S = -1+1 = 0
Hence S = 0.
i.e. ab+a-b+1 =0

Note: Alternative solution: Best method:
We have $ab+a-b+1=\left( a-1 \right)\left( b+1 \right)+2\text{ (i)}$
Now, we have $a-1=\sec x-\tan x-1=\dfrac{1-\sin x-\cos x}{\cos x}$ and $b+1=\csc x+\cot x+1=\dfrac{1+\cos x+\sin x}{\sin x}$
Hence, we have $\left( a-1 \right)\left( b+1 \right)=\dfrac{1-\left( \sin x+\cos x \right)}{\cos x}\times \dfrac{1+\left( \sin x+\cos x \right)}{\sin x}$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Using the above algebraic identity, we get
$\left( a-1 \right)\left( b+1 \right)=\dfrac{1-{{\left( \sin x+\cos x \right)}^{2}}}{\sin x\cos x}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Using the above identity, we get
$\left( a-1 \right)\left( b+1 \right)=\dfrac{1-\left( {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x \right)}{\sin x\cos x}$
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
Using the above identity, we get
$\left( a-1 \right)\left( b+1 \right)=\dfrac{1-1-2\sin x\cos x}{\sin x\cos x}=-2$
Hence from equation (i), we have
$ab+a-b+1=\left( a-1 \right)\left( b+1 \right)+2=-2+2=0$
Hence, we have ab+a-b+1 = 0.