Question

If cosecX-sinX =a and secX-cosX=b then prove that


$(a^2b)^2^/^3 +(ab^2)^2^/^3=1$

Answer 1

Answer:

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Answer 2

Given:

• cosec(x)-sin(x) = a
• sec(x)-cos(x) = b

To Find:

To prove that $(a^{2}b)^{2/3}+(ab^{2}) ^{2/3}=1$

Solution:

• we are going to prove it by using trigonometric standard equations.

• First consider, cosec(x)-sin(x) = a
• $\frac{1}{sin(x)}-sin(x) = a$
• $\frac{1-sin^{2}(x) }{sin(x)} = a$
• $\frac{cos^{2}(x) }{sin(x)} = a$
• Next consider, sec(x)-cos(x)=b
• $\frac{1}{cos(x)}-cos(x) = b$

• $\frac{1-cos^{2}(x) }{cos(x)}=b$
• Substituting the values of a and b in the equation  $(a^{2}b)^{2/3}+(ab^{2}) ^{2/3}=1$
• We get, $[(\frac{cos^{2}(x) }{sin(x)} )^{2} *\frac{sin^{2}(x) }{cos(x)} ]^{2/3} +[(\frac{cos^{2}(x) }{sin(x)} )*(\frac{sin^{2}(x) }{cos(x)} )^{2}]^{2/3}$
• $[(\frac{cos^{4}(x) }{sin^{2} (x)} )*\frac{sin^{2}(x) }{cos(x)} ]^{2/3} +[(\frac{cos^{2}(x) }{sin(x)} )*(\frac{sin^{4}(x) }{cos^{2} (x)}]^{2/3}$  
• In the above step, we are canceling the like terms and making the equation simple for us to solve.
• $cos^{2}(x)+sin^{2}(x)$ = 1(Standard trigonometry formula)

Hence Proved. $(a^{2}b)^{2/3}+(ab^{2}) ^{2/3}=1$