Question

Prove the following identities:
1) \[({\sin ^8}\theta - {\cos ^8}\theta ) = ({\sin ^2}\theta - {\cos ^2}\theta )(1 - 2{\sin ^2}\theta {\cos ^2}\theta )\]
2) \[{(1 + \tan A\tan B)^2} + {(\tan A - \tan B)^2} = {\sec ^2}A{\sec ^2}B\]
3) \[{(\tan A + \cos ecB)^2} - {(\cot B - \sec A)^2} = 2\tan A\cot B(\cos ecA + \sec B)\]

Answer 1

Hint:
(1) We prove the first identity using the concept of breaking the powers of 8 in a way that they open up as square of first term minus square of second term which can be opened with help of \[(a - b)(a + b) = {a^2} - {b^2}\].
 Also, we know \[{a^{mn}} = {({a^m})^n}\]
(2) We prove this identity by general opening of brackets using the formula \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] and then substitute \[1 + {\tan ^2}\theta = {\sec ^2}\theta \] wherever required.
(3) We prove this identity by general opening of brackets using the formula \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] and then substitute \[1 + {\tan ^2}\theta = {\sec ^2}\theta \] and \[1 + {\cot ^2}\theta = \cos e{c^2}\theta \] wherever required.

Complete step-by-step answer:
(1) \[({\sin ^8}\theta - {\cos ^8}\theta ) = ({\sin ^2}\theta - {\cos ^2}\theta )(1 - 2{\sin ^2}\theta {\cos ^2}\theta )\]
We can write \[8 = 4 \times 2\]
Therefore, LHS of the equation becomes
\[ \Rightarrow {({\sin ^4}\theta )^2} - {({\cos ^4}\theta )^2}\]
We know that \[(a - b)(a + b) = {a^2} - {b^2}\]
Substituting \[a = ({\sin ^4}\theta ), b = ({\cos ^4}\theta )\]
\[ \Rightarrow {({\sin ^4}\theta )^2} - {({\cos ^4}\theta )^2} = \left( {({{\sin }^4}\theta ) - ({{\cos }^4}\theta )} \right) \times \left( {({{\sin }^4}\theta ) + ({{\cos }^4}\theta )} \right)\]
We can write \[4 = 2 \times 2\]
We know that \[(a - b)(a + b) = {a^2} - {b^2}\]
Substituting \[a = ({\sin ^2}\theta ), b = ({\cos ^2}\theta )\] in the first bracket and breaking the powers in the second bracket.
\[ \Rightarrow {({\sin ^4}\theta )^2} - {({\cos ^4}\theta )^2} = \left( {({{\sin }^2}\theta - {{\cos }^2}\theta ) \times ({{\sin }^2}\theta + {{\cos }^2}\theta )} \right) \times \left( {{{({{\sin }^2}\theta )}^2} + {{({{\cos }^2}\theta )}^2}} \right)\]
Substitute the value of \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] in the equation.
\[ \Rightarrow {({\sin ^4}\theta )^2} - {({\cos ^4}\theta )^2} = \left( {{{\sin }^2}\theta - {{\cos }^2}\theta } \right) \times \left( {{{({{\sin }^2}\theta )}^2} + {{({{\cos }^2}\theta )}^2}} \right)\] … (i)
Now we know \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
Substituting \[a = ({\sin ^2}\theta ),b = ({\cos ^2}\theta )\]
\[ \Rightarrow {({\sin ^2}\theta + {\cos ^2}\theta )^2} = {({\sin ^2}\theta )^2} + {({\cos ^2}\theta )^2} + 2{\sin ^2}\theta {\cos ^2}\theta \]
We can write by shifting the terms
\[ \Rightarrow {({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}\theta = {({\sin ^2}\theta )^2} + {({\cos ^2}\theta )^2}\]
Substituting the value of \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[
   \Rightarrow {(1)^2} - 2{\sin ^2}\theta {\cos ^2}\theta = {({\sin ^2}\theta )^2} + {({\cos ^2}\theta )^2} \\
   \Rightarrow 1 - 2{\sin ^2}\theta {\cos ^2}\theta = {({\sin ^2}\theta )^2} + {({\cos ^2}\theta )^2} \\
 \] … (ii)
Substituting the value from equation (ii) in equation (i)
\[ \Rightarrow {({\sin ^4}\theta )^2} - {({\cos ^4}\theta )^2} = \left( {{{\sin }^2}\theta - {{\cos }^2}\theta } \right) \times \left( {1 - 2{{\sin }^2}\theta {{\cos }^2}\theta } \right)\]
This is equal to the LHS of the equation.
Hence Proved.
(2) \[{(1 + \tan A\tan B)^2} + {(\tan A - \tan B)^2} = {\sec ^2}A{\sec ^2}B\]
Consider the LHS of the equation.
Since, we know \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
Substituting \[a = 1, b = \tan A\tan B\]
\[ \Rightarrow {(1 + \tan A\tan B)^2} = 1 + {\tan ^2}A{\tan ^2}B + 2\tan A\tan B\] … (iii)
Since, we know \[{(a - b)^2} = {a^2} + {b^2} - 2ab\]
Substituting \[a = \tan A, b = \tan B\] in the formula
\[ \Rightarrow {(\tan A - \tan B)^2} = {\tan ^2}A + {\tan ^2}B - 2\tan A\tan B\] … (iv)
Substituting values from equation (iii) and (iv) in LHS we get
\[ \Rightarrow 1 + {\tan ^2}A{\tan ^2}B + 2\tan A\tan B + {\tan ^2}A + {\tan ^2}B - 2\tan A\tan B\]
Cancel out the terms with opposite sign
\[
   \Rightarrow 1 + {\tan ^2}A{\tan ^2}B + {\tan ^2}A + {\tan ^2}B \\
   \Rightarrow 1 + {\tan ^2}A + {\tan ^2}B + {\tan ^2}A{\tan ^2}B \\
 \]
Now we group together \[\left( {1 + {{\tan }^2}A} \right)\] as first part and make factors by taking common \[\left( {{{\tan }^2}B} \right)\] in second part
\[
   \Rightarrow (1 + {\tan ^2}A) + {\tan ^2}B(1 + {\tan ^2}A) \\
   \Rightarrow (1 + {\tan ^2}A)(1 + {\tan ^2}B) \\
 \]
Since we know \[1 + {\tan ^2}\theta = {\sec ^2}\theta \]
\[ \Rightarrow {\sec ^2}A{\sec ^2}B\]
This is equal to the RHS of the equation.
Hence Proved.
(3) \[{(\tan A + \cos ecB)^2} - {(\cot B - \sec A)^2} = 2\tan A\cot B(\cos ecA + \sec B)\]
Consider the LHS of the equation.
Since, we know \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
Substituting \[a = \tan A, b = \cos ecB\]
\[ \Rightarrow {(\tan A + \cos ecB)^2} = {\tan ^2}A + \cos e{c^2}B + 2\tan A\cos ecB\] … (iii)
Now substituting \[a = \cot B, b = \sec A\] in the formula \[{(a - b)^2} = {a^2} + {b^2} - 2ab\]
\[ \Rightarrow {(\cot B - \sec A)^2} = {\cot ^2}B + {\sec ^2}A - 2\cot B\sec A\] … (iv)
Substituting values from equation (iii) and (iv) in LHS we get
\[
   \Rightarrow ({\tan ^2}A + \cos e{c^2}B + 2\tan A\cos ecB) - ({\cot ^2}B + {\sec ^2}A - 2\cot B\sec A) \\
   \Rightarrow {\tan ^2}A + \cos e{c^2}B + 2\tan A\cos ecB - {\cot ^2}B - {\sec ^2}A + 2\cot B\sec A \\
   \Rightarrow {\tan ^2}A - {\sec ^2}A + \cos e{c^2}B - {\cot ^2}B + 2\tan A\cos ecB + 2\cot B\sec A \\
 \]
Substitute the values from the identities \[\left( {{{\tan }^2}\theta = {{\sec }^2}\theta - 1} \right)\] and \[\left( {1 + {{\cot }^2}\theta = \cos e{c^2}\theta } \right)\]
\[
   \Rightarrow ({\sec ^2}A - 1) - {\sec ^2}A + (1 + {\cot ^2}B) - {\cot ^2}B + 2\tan A\cos ecB + 2\cot B\sec A \\
   \Rightarrow - 1 + 1 + 2\tan A\cos ecB + 2\cot B\sec A \\
   \Rightarrow 2\tan A\cos ecB + 2\cot B\sec A \\
 \]
Now we divide and multiply first term by \[\left( {\cos B} \right)\] and divide and multiply second term by \[\left( {\sin A} \right)\]
\[ \Rightarrow 2\tan A\cos ecB\dfrac{{\cos B}}{{\cos B}} + 2\cot B\sec A\dfrac{{\sin A}}{{\sin A}}\]
Now we can write \[\left( {\cos ecB = \dfrac{1}{{\sin B}}} \right),\left( {\sec A = \dfrac{1}{{\cos A}}} \right)\]
\[
   \Rightarrow 2\tan A\dfrac{1}{{\sin B}}\dfrac{{\cos B}}{{\cos B}} + 2\cot B\dfrac{1}{{\cos A}}\dfrac{{\sin A}}{{\sin A}} \\
   \Rightarrow 2\tan A\dfrac{{\cos B}}{{\sin B}}\dfrac{1}{{\cos B}} + 2\cot B\dfrac{{\sin A}}{{\cos A}}\dfrac{1}{{\sin A}} \\
 \]
Now write \[\left( {\dfrac{{\cos B}}{{\sin B}} = \cot B} \right),\left( {\dfrac{1}{{\cos B}} = \sec B} \right),\left( {\dfrac{{\sin A}}{{\cos A}} = \tan A} \right),\left( {\dfrac{1}{{\sin A}} = \cos ecA} \right)\]
\[
   \Rightarrow 2\tan A\cot B\sec B + 2\cot B\tan A\cos ecA \\
   \Rightarrow 2\tan A\cot B(\sec B + \cos ecA) \\
 \]
This is equal to the LHS of the equation.
Hence Proved.

Note: Students many times make the mistake of substituting wrong values in the formula, they should always do substitution of angle and then open up the identity according to the formula.