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Question: Prove the following identities: 1) \[({\sin ^8}\theta - {\cos ^8}\theta ) = ({\sin ^2}\theta - {\c...

Prove the following identities:

  1. (sin⁡8θ−cos⁡8θ)=(sin⁡2θ−cos⁡2θ)(1−2sin⁡2θcos⁡2θ)({\sin ^8}\theta - {\cos ^8}\theta ) = ({\sin ^2}\theta - {\cos ^2}\theta )(1 - 2{\sin ^2}\theta {\cos ^2}\theta )
  2. (1+tan⁡Atan⁡B)2+(tan⁡A−tan⁡B)2=sec⁡2Asec⁡2B{(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=2tan⁡Acot⁡B(cos⁡ecA+sec⁡B){(\tan A + \cos ecB)^2} - {(\cot B - \sec A)^2} = 2\tan A\cot B(\cos ecA + \sec B)
Explanation

Solution

(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)=a2−b2(a - b)(a + b) = {a^2} - {b^2}.
Also, we know amn=(am)n{a^{mn}} = {({a^m})^n}
(2) We prove this identity by general opening of brackets using the formula (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab and then substitute 1+tan⁡2θ=sec⁡2θ1 + {\tan ^2}\theta = {\sec ^2}\theta wherever required.
(3) We prove this identity by general opening of brackets using the formula (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab and then substitute 1+tan⁡2θ=sec⁡2θ1 + {\tan ^2}\theta = {\sec ^2}\theta and 1+cot⁡2θ=cos⁡ec2θ1 + {\cot ^2}\theta = \cos e{c^2}\theta wherever required.

Complete step-by-step answer:
(1) (sin⁡8θ−cos⁡8θ)=(sin⁡2θ−cos⁡2θ)(1−2sin⁡2θcos⁡2θ)({\sin ^8}\theta - {\cos ^8}\theta ) = ({\sin ^2}\theta - {\cos ^2}\theta )(1 - 2{\sin ^2}\theta {\cos ^2}\theta )
We can write 8=4×28 = 4 \times 2
Therefore, LHS of the equation becomes
⇒(sin⁡4θ)2−(cos⁡4θ)2\Rightarrow {({\sin ^4}\theta )^2} - {({\cos ^4}\theta )^2}
We know that (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}
Substituting a=(sin⁡4θ),b=(cos⁡4θ)a = ({\sin ^4}\theta ), b = ({\cos ^4}\theta )
⇒(sin⁡4θ)2−(cos⁡4θ)2=((sin⁡4θ)−(cos⁡4θ))×((sin⁡4θ)+(cos⁡4θ))\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=2×24 = 2 \times 2
We know that (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}
Substituting a=(sin⁡2θ),b=(cos⁡2θ)a = ({\sin ^2}\theta ), b = ({\cos ^2}\theta ) in the first bracket and breaking the powers in the second bracket.
⇒(sin⁡4θ)2−(cos⁡4θ)2=((sin⁡2θ−cos⁡2θ)×(sin⁡2θ+cos⁡2θ))×((sin⁡2θ)2+(cos⁡2θ)2)\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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 in the equation.
⇒(sin⁡4θ)2−(cos⁡4θ)2=(sin⁡2θ−cos⁡2θ)×((sin⁡2θ)2+(cos⁡2θ)2)\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=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
Substituting a=(sin⁡2θ),b=(cos⁡2θ)a = ({\sin ^2}\theta ),b = ({\cos ^2}\theta )
⇒(sin⁡2θ+cos⁡2θ)2=(sin⁡2θ)2+(cos⁡2θ)2+2sin⁡2θcos⁡2θ\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
⇒(sin⁡2θ+cos⁡2θ)2−2sin⁡2θcos⁡2θ=(sin⁡2θ)2+(cos⁡2θ)2\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⁡2θ+cos⁡2θ=1{\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) \\

ThisisequaltotheLHSoftheequation.HenceProved.∗∗Note:∗∗Studentsmanytimesmakethemistakeofsubstitutingwrongvaluesintheformula,theyshouldalwaysdosubstitutionofangleandthenopenuptheidentityaccordingtotheformula.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.