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
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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.
(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.
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