Solveeit Logo
Login

Question

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

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

Explanation

Solution

Hint: Convert the expression ab+a−b+1ab+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⁡2x+cos⁡2x=1{{\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=1cos⁡x,csc⁡x=1sin⁡x,cot⁡x=cos⁡xsin⁡x\sec x=\dfrac{1}{\cos x},\csc x=\dfrac{1}{\sin x},\cot x=\dfrac{\cos x}{\sin x} and tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}, we get
S =(1cos⁡x−sin⁡xcos⁡x)(1sin⁡x+cos⁡xsin⁡x)+(1cos⁡x−sin⁡xcos⁡x)−(1sin⁡x+cos⁡xsin⁡x)+1=\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 =1−sin⁡xcos⁡x×1+cos⁡xsin⁡x+1−sin⁡xcos⁡x−1+cos⁡xsin⁡x+1=\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 =(1−sin⁡x)(1+cos⁡x)+(1−sin⁡x)sin⁡x−(1+cos⁡x)cos⁡xsin⁡xcos⁡x+1=\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 =1−sin⁡x+cos⁡x−cos⁡xsin⁡x+sin⁡x−sin⁡2x−cos⁡x−cos⁡2xsin⁡xcos⁡x+1=\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⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1.
Using the above identity, we get
S =1−1−cos⁡xsin⁡xcos⁡xsin⁡x+1=\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=(a−1)(b+1)+2 (i)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=1−sin⁡x−cos⁡xcos⁡xa-1=\sec x-\tan x-1=\dfrac{1-\sin x-\cos x}{\cos x} and b+1=csc⁡x+cot⁡x+1=1+cos⁡x+sin⁡xsin⁡xb+1=\csc x+\cot x+1=\dfrac{1+\cos x+\sin x}{\sin x}
Hence, we have (a−1)(b+1)=1−(sin⁡x+cos⁡x)cos⁡x×1+(sin⁡x+cos⁡x)sin⁡x\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 (a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}
Using the above algebraic identity, we get
(a−1)(b+1)=1−(sin⁡x+cos⁡x)2sin⁡xcos⁡x\left( a-1 \right)\left( b+1 \right)=\dfrac{1-{{\left( \sin x+\cos x \right)}^{2}}}{\sin x\cos x}
We know that (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}
Using the above identity, we get
(a−1)(b+1)=1−(sin⁡2x+cos⁡2x+2sin⁡xcos⁡x)sin⁡xcos⁡x\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⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1.
Using the above identity, we get
(a−1)(b+1)=1−1−2sin⁡xcos⁡xsin⁡xcos⁡x=−2\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=(a−1)(b+1)+2=−2+2=0ab+a-b+1=\left( a-1 \right)\left( b+1 \right)+2=-2+2=0
Hence, we have ab+a-b+1 = 0.