In trigonometry chapter of ncert class 10 what is the best solution of this question please help me to solve this tricky question in simply or easy way of exercise 8.4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A .

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# Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A .Q.5(8)

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(sin A + cosec A)

^{2 }+ (cos A + sec A)^{2}= 7+tan^{2}A+cot^{2}AL.H.S. = (sin A + cosec A)

^{2 }+ (cos A + sec A)^{2}It is of the form (a+b)

^{2}, expand it(a+b)

^{2}=a^{2}+ b^{2}+2ab^{ }= (sin^{2}A + cosec^{2}A + 2 sin A cosec A) + (cos^{2}A + sec^{2}A + 2 cos A sec A)= (sin

^{2}A + cos^{2}A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan^{2}A + 1 + cot^{2}A= 1 + 2 + 2 + 2 + tan

^{2}A + cot^{2}A= 7+tan

^{2}A+cot^{2}A = R.H.S.Therefore, (sin A + cosec A)

^{2 }+ (cos A + sec A)^{2}= 7+tan^{2}A+cot^{2}AHence proved.