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Question: Prove the following identities, where the angles involved are acute angles for which the expressions...

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosecθ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ{\left( {cosec\theta - \cot \theta } \right)^2} = \dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}
(ii)cos⁡A1+sin⁡A+1+sin⁡Acos⁡A\dfrac{{\cos A}}{{1 + \sin A}} + \dfrac{{1 + \sin A}}{{\cos A}} =sec⁡A = \sec A
(iii) tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θ⋅cosecθ\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta \cdot cosec\theta
(iv) 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\dfrac{{1 + \sec A}}{{\sec A}} = \dfrac{{{{\sin }^2}A}}{{1 - \cos A}}
(v) cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cos⁡ecA+cot⁡A\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A
(vi) 1+sin⁡A1−sin⁡A=sec⁡A+tan⁡A\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} = \sec A + \tan A
(vii) sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3}\theta - \cos \theta }} = \tan \theta
(viii) (sin⁡A+cos⁡ecA)2+(cos⁡A+sec⁡A)2=7+tan⁡2A+cot⁡2A{\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2} = 7 + {\tan ^2}A + {\cot ^2}A
(ix) (cosecA−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A\left( {cosecA - \sin A} \right)\left( {\sec A - \cos A} \right) = \dfrac{1}{{\tan A + \cot A}}
(x) [1+tan⁡2A1+cot⁡2A]=[1−tan⁡A1−citA]2=tan⁡2A\left[ {\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}} \right] = {\left[ {\dfrac{{1 - \tan A}}{{1 - citA}}} \right]^2} = {\tan ^2}A

Explanation

Solution

Hint: We will prove in all the parts of the question that the Left Hand Side and the Right Hand Side of the trigonometric equation are equal, for which, we first write down the given equation and start solving either the LHS or the RHS of the equation by taking the help of various trigonometric identities.

Complete step-by-step answer:

(i) (cosecθ−cot⁡θ)2=1−cot⁡θ1+cot⁡θ{\left( {cosec\theta - \cot \theta } \right)^2} = \dfrac{{1 - \cot \theta }}{{1 + \cot \theta }}
Taking LHS, we get,
(cosecθ−cot⁡θ)2{\left( {cosec\theta - \cot \theta } \right)^2}……. (1)\left( 1 \right)
Now, we will make it in terms of sin⁡θandcos⁡θ\sin \theta and\cos \theta .
As we know that, cosecθ=1sin⁡θcosec\theta = \dfrac{1}{{\sin \theta }}and cot⁡θ=cos⁡θsin⁡θ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}
Therefore, equation (1) becomes,
= (1sin⁡θ−cos⁡θsin⁡θ)2\left( {\dfrac{1}{{\sin \theta }}} \right. - {\left. {\dfrac{{\cos \theta }}{{\sin \theta }}} \right)^2}
= (1−cos⁡θsin⁡θ)2{\left( {\dfrac{{1 - \cos \theta }}{{\sin \theta }}} \right)^2}
= (1−cos⁡θ)2sin⁡2θ\dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{{{\sin }^2}\theta }}
We know that, cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1
\Rightarrow $$$${\sin ^2}\theta = 1 - {\cos ^2}\theta
Therefore, (1−cos⁡θ)21−cos⁡2θ\dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{1 - {{\cos }^2}\theta }}
= (1−cos⁡θ)212−cos⁡2θ\dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{{1^2} - {{\cos }^2}\theta }}
Since, in the denominator , 12−cos⁡2θ{1^2} - {\cos ^2}\theta represents an identity, i.e., {a^2} - {b^2}$$$$ = \left( {a - b} \right)\left( {a + b} \right),then it becomes,
= (1−cos⁡θ)2(1+cos⁡θ)(1−cos⁡θ)\dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}}
= (1−cos⁡θ)(1−cos⁡θ)(1+cos⁡θ)(1−cos⁡θ)\dfrac{{\left( {1 - \cos \theta } \right)\left( {1 - \cos \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}}
= 1−cos⁡θ1+cos⁡θ\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}
=RHS
Hence Proved.

(ii) \dfrac{{\cos A}}{{1 + \sin A}} + \dfrac{{1 + \sin A}}{{\cos A}}$$$$ = \sec A
Taking LHS, we get
cos⁡A1+sin⁡A+1+sin⁡Acos⁡A\dfrac{{\cos A}}{{1 + \sin A}} + \dfrac{{1 + \sin A}}{{\cos A}} , which is already in the terms of sine and cosine
Now, by taking LCM, we get
=cos⁡2A+(1+sin⁡A)2cos⁡A(1+sin⁡A)\dfrac{{{{\cos }^2}A + {{\left( {1 + \sin A} \right)}^2}}}{{\cos A\left( {1 + \sin A} \right)}} , where in the numerator, (1+sin⁡A)2{\left( {1 + \sin A} \right)^2}represents an identity, i.e., (a+b)2{\left( {a + b} \right)^2},then it becomes,
= cos⁡2A+1+sin⁡2A+2sin⁡Acos⁡A(1+sin⁡A)\dfrac{{{{\cos }^2}A + 1 + {{\sin }^2}A + 2\sin A}}{{\cos A\left( {1 + \sin A} \right)}}
As we know that, cos⁡2A+sin⁡2A=1{\cos ^2}A + {\sin ^2}A = 1,therefore, we get
= 1+1+2sin⁡Acos⁡A(1+sin⁡A)\dfrac{{1 + 1 + 2\sin A}}{{\cos A\left( {1 + \sin A} \right)}}
= 2+2sin⁡Acos⁡A(1+sin⁡A)\dfrac{{2 + 2\sin A}}{{\cos A\left( {1 + \sin A} \right)}}
Now, taking 2 common in the numerator, we get
=2(1+sin⁡A)cos⁡A(1+sinA⁡)\dfrac{{2\left( {1 + \sin A} \right)}}{{\cos A\left( {1 + \operatorname{sinA} } \right)}}
By cutting equal terms in the numerator and denominator, we get
=2cos⁡A\dfrac{2}{{\cos A}}
=2⋅1cos⁡A2 \cdot \dfrac{1}{{\cos A}}
=sec⁡A\sec A
=RHS
Hence Proved.

(iii) tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θcos⁡ecθ\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta \cos ec\theta
Taking LHS, we get
tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }}

Now, we will make it in terms of sin⁡θandcos⁡θ\sin \theta and\cos \theta .
We know that, tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} and cot⁡θ=cos⁡θsin⁡θ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}, then it becomes,
= sin⁡θcos⁡θ1−cos⁡θsin⁡θ+cos⁡θsin⁡θ1−sin⁡θcos⁡θ\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{1 - \dfrac{{\cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin \theta }}}}{{1 - \dfrac{{\sin \theta }}{{\cos \theta }}}}
Now, by taking LCM in the denominator, we get

=sin⁡θcos⁡θsin⁡θ−cos⁡θsin⁡θ+cos⁡θsin⁡cos⁡θ−sin⁡θcos⁡θ\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin }}}}{{\dfrac{{\cos \theta - \sin \theta }}{{\cos \theta }}}}
By solving, it becomes
= sin⁡θcos⁡θ×sin⁡θsin⁡θ−cos⁡θ+cos⁡θsin⁡θ×cos⁡θcos⁡θ−sin⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} \times \dfrac{{\sin \theta }}{{\sin \theta - \cos \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }} \times \dfrac{{\cos \theta }}{{\cos \theta - \sin \theta }}

=sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)+cos⁡2θsin⁡θ(cos⁡θ−sin⁡θ)\dfrac{{{{\sin }^2}\theta }}{{\cos \theta \left( {\sin \theta - \cos \theta } \right)}} + \dfrac{{{{\cos }^2}\theta }}{{\sin \theta \left( {\cos \theta - \sin \theta } \right)}}
By taking minus common from the second term,
= sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)−cos⁡2θsin⁡θ(sin⁡θ−cos⁡θ)\dfrac{{{{\sin }^2}\theta }}{{\cos \theta \left( {\sin \theta - \cos \theta } \right)}} - \dfrac{{{{\cos }^2}\theta }}{{\sin \theta \left( {\sin \theta - \cos \theta } \right)}}

= sin⁡3θ−cos⁡3θcos⁡θ⋅sin⁡θ(sin⁡θ−cos⁡θ)\dfrac{{{{\sin }^3}\theta - {{\cos }^3}\theta }}{{\cos \theta \cdot \sin \theta \left( {\sin \theta - \cos \theta } \right)}}
Now, in the numerator, sin⁡3θ−cos⁡3θ{\sin ^3}\theta - {\cos ^3}\theta represents an identity, i.e., a3−b3=(a−b)(a2+ab+b2){a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right), therefore, it becomes,
= (sin⁡θ−cos⁡θ)(sin⁡2θ+sin⁡θ⋅cos⁡θ+cos⁡2θ)cos⁡θ⋅sin⁡θ(sin⁡θ−cos⁡θ)\dfrac{{\left( {\sin \theta - \cos \theta } \right)\left( {{{\sin }^2}\theta + \sin \theta \cdot \cos \theta + {{\cos }^2}\theta } \right)}}{{\cos \theta \cdot \sin \theta \left( {\sin \theta - \cos \theta } \right)}}
So, by cutting the equal terms in the numerator and denominator, we obtain
=(sin⁡2θ+cos⁡2θ+sin⁡θ⋅cos⁡θ)cos⁡θ⋅sin⁡θ\dfrac{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta + \sin \theta \cdot \cos \theta } \right)}}{{\cos \theta \cdot \sin \theta }}
= 1+sin⁡θ⋅cos⁡θcos⁡θ⋅sin⁡θ\dfrac{{1 + \sin \theta \cdot \cos \theta }}{{\cos \theta \cdot \sin \theta }} [∵sin⁡2θ+cos⁡2θ=1]\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]
Now, by separating the terms, we obtain
= 1cos⁡θ⋅sin⁡θ+sin⁡θ⋅cos⁡θsinθ⋅cos⁡θ\dfrac{1}{{\cos \theta \cdot \sin \theta }} + \dfrac{{\sin \theta \cdot \cos \theta }}{{sin\theta \cdot \cos \theta }}
By cutting out the equal terms,
= sec⁡θ⋅cosecθ+1\sec \theta \cdot cosec\theta + 1 , which can also be written as,
=1+sec⁡θ⋅cosecθ1 + \sec \theta \cdot cosec\theta
=RHS
Hence Proved.

(iv) 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\dfrac{{1 + \sec A}}{{\sec A}} = \dfrac{{{{\sin }^2}A}}{{1 - \cos A}}
Taking LHS, we get
1+sec⁡Asec⁡A\dfrac{{1 + \sec A}}{{\sec A}}
Now, we will make it in terms of cosine.
As we know that, sec⁡A=1cos⁡A\sec A = \dfrac{1}{{\cos A}}, then,
= $$\dfrac{{1 + \dfrac{1}{{\cos A}}}}{{\dfrac{1}{{\cos A}}}}$$ = $$\dfrac{{\dfrac{{\cos A + 1}}{{\cos A}}}}{{\dfrac{1}{{\cos A}}}}$$ Now, by evaluating it, we obtain = $$\dfrac{{\cos A + 1}}{{\cos A}} \times \dfrac{{\cos A}}{1}$$ By cutting out equal terms, we get =$$\cos A + 1$$ Or $$1 + \cos A$$ Now, taking RHS, we get $$\dfrac{{{{\sin }^2}A}}{{1 - \cos A}}$$ We know that, $${\sin ^2}A + {\cos ^2}A = 1$$ $$ \Rightarrow {\sin ^2}A = 1 - {\cos ^2}A$$ Therefore, it becomes = $$\dfrac{{1 - {{\cos }^2}A}}{{1 - \operatorname{cosA} }}$$ = $$\dfrac{{{{\left( 1 \right)}^2} - {{\left( {\cos A} \right)}^2}}}{{1 - \cos A}}$$ Now, in the numerator, $${\left( 1 \right)^2} - {\left( {\cos A} \right)^2}$$ represents an identity, i.e., $${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right),thenitbecomes,=, then it becomes, = \dfrac{{\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)}}{{\left( {1 - \cos A} \right)}}Bycuttingequalterms,weobtain= By cutting equal terms, we obtain =1 + \cos A \therefore $$ LHS=RHS
Hence Proved.

(v) cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cos⁡ecA+cot⁡A\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A
Taking LHS, we get
cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}}
Now, dividing each term of numerator and denominator by sin⁡A\sin A, we obtain
= cos⁡Asin⁡A−sin⁡Asin⁡A+1sin⁡Acos⁡Asin⁡A+sin⁡Asin⁡A−1sin⁡A\dfrac{{\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} - \dfrac{1}{{\sin A}}}}
Cutting out all the equal terms, we get
= cos⁡Asin⁡A−1+1sin⁡Acos⁡Asin⁡A+1−1sin⁡A\dfrac{{\dfrac{{\cos A}}{{\sin A}} - 1 + \dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + 1 - \dfrac{1}{{\sin A}}}}
We know that, cos⁡Asin⁡A=cot⁡A\dfrac{{\cos A}}{{\sin A}} = \cot A and 1sin⁡A=cos⁡ecA\dfrac{1}{{\sin A}} = \cos ecA, then it becomes
= cotA−1+cosecAcot⁡A+1−cos⁡ecA\dfrac{{cotA - 1 + cosecA}}{{\cot A + 1 - \cos ecA}}
= (cos⁡ecA+cot⁡A)−1cot⁡A−cos⁡ecA+1\dfrac{{\left( {\cos ecA + \cot A} \right) - 1}}{{\cot A - \cos ecA + 1}}
= (cos⁡ecA+cot⁡A)−11−cos⁡ecA+cot⁡A\dfrac{{\left( {\cos ecA + \cot A} \right) - 1}}{{1 - \cos ecA + \cot A}}
As we know that, cos⁡ec2A−cot⁡2A=1\cos e{c^2}A - {\cot ^2}A = 1, therefore
= (cos⁡ecA+cot⁡A)−(cos⁡ec2A−cot⁡2A)1−cos⁡ecA+cot⁡A\dfrac{{\left( {\cos ecA + \cot A} \right) - \left( {\cos e{c^2}A - {{\cot }^2}A} \right)}}{{1 - \cos ecA + \cot A}}
We can write the above equation as
(cos⁡ecA+cot⁡A)−(cos⁡ecA−cot⁡A)(cos⁡ecA+cot⁡A)1−cos⁡ecA+cot⁡A\dfrac{{\left( {\cos ecA + \cot A} \right) - \left( {\cos ecA - \cot A} \right)\left( {\cos ecA + \cot A} \right)}}{{1 - \cos ecA + \cot A}} by using the identity {a^2} - {b^2}$$$$ = \left( {a - b} \right)\left( {a + b} \right)
Now, taking cos⁡ecA+cot⁡A\cos ecA + \cot A common in the numerator, we get
= (cos⁡ecA+cot⁡A)[1−(cos⁡ecA−cot⁡A)]1−cos⁡ecA+cot⁡A\dfrac{{\left( {\cos ecA + \cot A} \right)\left[ {1 - \left( {\cos ecA - \cot A} \right)} \right]}}{{1 - \cos ecA + \cot A}}
= (cos⁡ecA+cot⁡A)(1−cos⁡ecA+cot⁡A)(1−cos⁡ecA+cot⁡A)\dfrac{{\left( {\cos ecA + \cot A} \right)\left( {1 - \cos ecA + \cot A} \right)}}{{\left( {1 - \cos ecA + \cot A} \right)}}
Now, by cutting out the equal terms,
= cos⁡ecA+cot⁡A\cos ecA + \cot A
=RHS
Hence Proved.

(vi) 1+sin⁡A1−sin⁡A=sec⁡A+tanA⁡\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} = \sec A + \operatorname{tanA}
Taking LHS, we get
1+sin⁡A1−sin⁡A\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}}
Using rationalization, we obtain
= 1+sin⁡A1−sin⁡A×1+sin⁡A1+sin⁡A\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}} \times \dfrac{{1 + \sin A}}{{1 + \sin A}}}
= (1+sin⁡A)2(1−sin⁡A)(1+sin⁡A)\sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{\left( {1 - \sin A} \right)\left( {1 + \sin A} \right)}}}
= (1+sin⁡A)2(1)2−(sin⁡A)2\sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{{{\left( 1 \right)}^2} - {{\left( {\sin A} \right)}^2}}}} [∵(a−b)(a+b)=a2−b2]\left[ {\because \left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}} \right]

= (1+sin⁡A)21−sin⁡2A\sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{1 - {{\sin }^2}A}}}
We know that, sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1
⇒1−sin⁡2A=cos⁡2A\Rightarrow 1 - {\sin ^2}A = {\cos ^2}A
Therefore, it becomes
= (1+sin⁡A)2cos⁡2A\sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{{{\cos }^2}A}}}
= (1+sin⁡Acos⁡A)2\sqrt {{{\left( {\dfrac{{1 + \sin A}}{{\cos A}}} \right)}^2}}
Now, by cutting out the square and the square root, we obtain
= 1+sin⁡Acos⁡A\dfrac{{1 + \sin A}}{{\cos A}}
= 1cos⁡A+sin⁡Acos⁡A\dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}}
= sec⁡A+tan⁡A\sec A + \tan A [∵1cos⁡A=sec⁡Aandsin⁡Acos⁡A=tan⁡A]\left[ {\because \dfrac{1}{{\cos A}} = \sec Aand\dfrac{{\sin A}}{{\cos A}} = \tan A} \right]

=RHS
Hence Proved.

(vii) sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3}\theta - \cos \theta }} = \tan \theta
Taking LHS, we get
sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ\dfrac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3}\theta - \cos \theta }}
Now, taking sine common in the numerator and cosine common in the denominator, we obtain
= sin⁡θ(1−2sin⁡2θ)cos⁡θ(2cos⁡2θ−1)\dfrac{{\sin \theta \left( {1 - 2{{\sin }^2}\theta } \right)}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}
= sin⁡θ[1−2(sin⁡2θ)]cos⁡θ(2cos⁡2θ−1)\dfrac{{\sin \theta \left[ {1 - 2\left( {{{\sin }^2}\theta } \right)} \right]}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}
=sin⁡θ[1−2(1−cos⁡2θ)]cos⁡θ(2cos⁡2θ−1)\dfrac{{\sin \theta \left[ {1 - 2\left( {1 - {{\cos }^2}\theta } \right)} \right]}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}} [∵sin⁡2θ+cos⁡2θ=1 ⇒sin⁡2θ=1−cos⁡2θ ]\left[ \begin{gathered} \because {\sin ^2}\theta + {\cos ^2}\theta = 1 \\\ \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta \\\ \end{gathered} \right]

= sin⁡θ(1−2+2cos⁡2θ)cos⁡θ(2cos⁡2θ−1)\dfrac{{\sin \theta \left( {1 - 2 + 2{{\cos }^2}\theta } \right)}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}
= sin⁡θ(−1+2cos⁡2θ)cos⁡θ(2cos⁡2θ−1)\dfrac{{\sin \theta \left( { - 1 + 2{{\cos }^2}\theta } \right)}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}
= sin⁡θ(2cos⁡2θ−1)cos⁡θ(2cos⁡2θ−1)\dfrac{{\sin \theta \left( {2{{\cos }^2}\theta - 1} \right)}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}
By cutting out the equal terms, we get
= sin⁡θcos⁡θ\dfrac{{\sin \theta }}{{\cos \theta }}
As we know that, sin⁡θcos⁡θ=tan⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta , therefore it becomes
= tan⁡θ\tan \theta
=RHS
Hence Proved.

(viii) (sin⁡A+cos⁡ecA)2+(cos⁡A+secA⁡)2=7+tan⁡2A+cot⁡2A{\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \operatorname{secA} } \right)^2} = 7 + {\tan ^2}A + {\cot ^2}A
Taking LHS, we get
(sin⁡A+cos⁡ecA)2+(cos⁡A+sec⁡A)2{\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2} $$$$
Here, (sin⁡A+cos⁡ecA)2and(cos⁡A+sec⁡A)2{\left( {\sin A + \cos ecA} \right)^2}and{\left( {\cos A + \sec A} \right)^2} represents an identity, i.e., (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}, therefore it becomes
= sin⁡2A+2sin⁡A⋅cos⁡ecA+cos⁡ec2A+cos⁡2A+2cos⁡A⋅sec⁡A+sec⁡2A{\sin ^2}A + 2\sin A \cdot \cos ecA + \cos e{c^2}A + {\cos ^2}A + 2\cos A \cdot \sec A + {\sec ^2}A
Since, cos⁡ecA=1sin⁡A,sec⁡A=1cos⁡A,cos⁡ec2A=1+cot⁡2Aandsec⁡2A=1+tan⁡2A\cos ecA = \dfrac{1}{{\sin A}},\sec A = \dfrac{1}{{\cos A}},\cos e{c^2}A = 1 + {\cot ^2}Aand{\sec ^2}A = 1 + {\tan ^2}A, therefore it becomes,
= sin⁡2A+cos⁡2A+2sin⁡A⋅1sin⁡A+1+cot⁡2A+2cos⁡A⋅1cos⁡A+1+tan⁡2A{\sin ^2}A + {\cos ^2}A + 2\sin A \cdot \dfrac{1}{{\sin A}} + 1 + {\cot ^2}A + 2\cos A \cdot \dfrac{1}{{\cos A}} + 1 + {\tan ^2}A
Now, after cutting out the equal terms and by using the formula sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1, we get
= 1+2+1+cot⁡2A+2+1+tan⁡2A1 + 2 + 1 + {\cot ^2}A + 2 + 1 + {\tan ^2}A
= 7+tan⁡2A+cot⁡2A7 + {\tan ^2}A + {\cot ^2}A
=RHS
Hence Proved.

(ix) (cosecA+sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A\left( {cosecA + \sin A} \right)\left( {\sec A - \cos A} \right) = \dfrac{1}{{\tan A + \cot A}}
Taking LHS, we get
(cosecA−sin⁡A)(sec⁡A−cos⁡A)\left( {cosecA - \sin A} \right)\left( {\sec A - \cos A} \right)
Now, we will make it in terms of sine and cosine.
As we know that, cosecθ=1sin⁡θcosec\theta = \dfrac{1}{{\sin \theta }} and sec⁡A=1cos⁡A\sec A = \dfrac{1}{{\cos A}}, therefore it becomes
= (1sin⁡A−sin⁡A)(1cos⁡A−cos⁡A)\left( {\dfrac{1}{{\sin A}} - \sin A} \right)\left( {\dfrac{1}{{\cos A}} - \cos A} \right)
By taking LCM, we get
= (1−sin⁡2Asin⁡A)(1−cos⁡2Acos⁡A)\left( {\dfrac{{1 - {{\sin }^2}A}}{{\sin A}}} \right)\left( {\dfrac{{1 - {{\cos }^2}A}}{{\cos A}}} \right)
= cos⁡2Asin⁡A×sin⁡2Acos⁡A\dfrac{{{{\cos }^2}A}}{{\sin A}} \times \dfrac{{{{\sin }^2}A}}{{\cos A}} [∵sin⁡2A+cos⁡2A=1 ⇒1−sin⁡2A=cos⁡2A also,sin⁡2A+cos⁡2A=1 ⇒1−cos⁡2A=sin⁡2A ]\left[ \begin{gathered} \because {\sin ^2}A + {\cos ^2}A = 1 \\\ \Rightarrow 1 - {\sin ^2}A = {\cos ^2}A \\\ also,{\sin ^2}A + {\cos ^2}A = 1 \\\ \Rightarrow 1 - {\cos ^2}A = {\sin ^2}A \\\ \end{gathered} \right]

By cutting out the equal terms, we obtain,
= cos⁡A⋅sin⁡A\cos A \cdot \sin A

Now, taking RHS, we get
1tan⁡A+cot⁡A\dfrac{1}{{\tan A + \cot A}}
We know that, cot⁡A=1tan⁡A\cot A = \dfrac{1}{{\tan A}}, therefore we obtain
= 1tan⁡A+1tan⁡\dfrac{1}{{\tan A + \dfrac{1}{{\tan }}}}
By taking LCM in the denominator, we get
= 1tan⁡2A+1tan⁡A\dfrac{1}{{\dfrac{{{{\tan }^2}A + 1}}{{\tan A}}}}
As we know that, tan⁡2A+1=sec⁡2A{\tan ^2}A + 1 = {\sec ^2}A, then it becomes
= 1sec⁡2Atan⁡A\dfrac{1}{{\dfrac{{{{\sec }^2}A}}{{\tan A}}}}
Now, we obtain
= tan⁡Asec⁡2A\dfrac{{\tan A}}{{{{\sec }^2}A}}
= sin⁡Acos⁡A1cos⁡2A\dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{\dfrac{1}{{{{\cos }^2}A}}}} [∵tan⁡A=sin⁡Acos⁡Aandsec⁡2A=1cos⁡2A]\left[ {\because \tan A = \dfrac{{\sin A}}{{\cos A}}and{{\sec }^2}A = \dfrac{1}{{{{\cos }^2}A}}} \right]
= sin⁡Acos⁡A×cos⁡2A\dfrac{{\sin A}}{{\cos A}} \times {\cos ^2}A
= sin⁡A⋅cos⁡A\sin A \cdot \cos A
∴\therefore LHS=RHS
Hence Proved.

(x) [1+tan⁡2A1+cot⁡2A]=[1−tan⁡A1−cotA]2=tan⁡2A\left[ {\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}} \right] = {\left[ {\dfrac{{1 - \tan A}}{{1 - cotA}}} \right]^2} = {\tan ^2}A
Taking LHS, we get
[1+tan⁡2A1+cot⁡2A]\left[ {\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}} \right]
We will make it in terms of sine and cosine.
As we know that, 1+tan⁡2A=sec⁡2Aand1+cot⁡2A=cos⁡ec2A1 + {\tan ^2}A = {\sec ^2}Aand1 + {\cot ^2}A = \cos e{c^2}A, then it becomes
= sec⁡2Acos⁡ec2A\dfrac{{{{\sec }^2}A}}{{\cos e{c^2}A}}
= 1cos⁡2A1sin⁡2A\dfrac{{\dfrac{1}{{{{\cos }^2}A}}}}{{\dfrac{1}{{{{\sin }^2}A}}}} [∵sec⁡A=1cos⁡Aandcos⁡ecA=1sin⁡A]\left[ {\because \sec A = \dfrac{1}{{\cos A}}and\cos ecA = \dfrac{1}{{\sin A}}} \right]
= 1cos⁡2A×sin⁡2A1\dfrac{1}{{{{\cos }^2}A}} \times \dfrac{{{{\sin }^2}A}}{1}
= sin⁡2Acos⁡2A\dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}
= tan⁡2A{\tan ^2}A
Now, taking RHS, we get
[1−tan⁡A1−cot⁡A]2{\left[ {\dfrac{{1 - \tan A}}{{1 - \cot A}}} \right]^2}
We will make it in terms of sine and cosine.
= (1−sin⁡Acos⁡A1−cos⁡Asin⁡A)2{\left( {\dfrac{{1 - \dfrac{{\sin A}}{{\cos A}}}}{{1 - \dfrac{{\cos A}}{{\sin A}}}}} \right)^2} [∵tan⁡A=sin⁡Acos⁡Aandcot⁡A=cos⁡Asin⁡A]\left[ {\because \tan A = \dfrac{{\sin A}}{{\cos A}}and\cot A = \dfrac{{\cos A}}{{\sin A}}} \right]
Taking LCM in the numerator and the denominator, we obtain
= (cos⁡A−sin⁡Acos⁡Asin⁡A−cos⁡Asin⁡A)2{\left( {\dfrac{{\dfrac{{\cos A - \sin A}}{{\cos A}}}}{{\dfrac{{\sin A - \cos A}}{{\sin A}}}}} \right)^2}
By further solving it, we get
= (cos⁡A−sin⁡A)2cos⁡2A×sin⁡2A(sin⁡A−cos⁡A)2\dfrac{{{{\left( {\cos A - \sin A} \right)}^2}}}{{{{\cos }^2}A}} \times \dfrac{{{{\sin }^2}A}}{{{{\left( {\sin A - \cos A} \right)}^2}}}
= [−(sin⁡A−cos⁡A)]2cos⁡2A×sin⁡2A(sin⁡A−cos⁡A)2\dfrac{{{{\left[ { - \left( {\sin A - \cos A} \right)} \right]}^2}}}{{{{\cos }^2}A}} \times \dfrac{{{{\sin }^2}A}}{{{{\left( {\sin A - \cos A} \right)}^2}}}
Now, by cutting out the equal terms and as we know the square of negative sign is positive, we obtain
= sin⁡2Acos⁡2A\dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}
= tan⁡2A{\tan ^2}A
∴\therefore LHS=RHS
Hence Proved.

Note: One should be careful while doing such questions because in these questions we have to use different identities which might be confusing sometimes and before doing these questions one must be clear headed in all aspects. These all parts are quite typical, so one must prove these parts as we did in the solution.