Prove the following identity, where the angles involved are acute angles for which the expression is defined.
$(s i n A + c o s e c A)^{2} + (c o s A + s e c A)^{2} = 7 + t a n^{2} A + c o t^{2} A$

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Solution

$(s i n A + c o s e c A)^{2} + (c o s A + s e c A)^{2} = 7 + t a n^{2} A + c o t^{2} A L . H . S = (s i n A + c o s e c A)^{2} + (c o s A + s e c A)^{2} = (s i n^{2} A + c o s e c^{2} A + 2 s i n A c o s e c A) + (c o s^{2} A + s e c^{2} A + 2 c o s A s e c A) = (s i n^{2} A + c o s^{2} A) + 2 s i n A (\frac{1}{s i n A}) + 2 c o s A (\frac{1}{c o s A}) + 1 + t a n^{2} A + 1 + c o t^{2} A = 1 + 2 + 2 + 2 + t a n^{2} A + c o t^{2} A = 7 + t a n^{2} A + c o t^{2} A = R . H . S$

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Q. Question 5 (viii)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
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(s i n A + c o s e c A)^{2} + (c o s A + s e c A)^{2} = 7 + t a n^{2} A + c o t^{2} A

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(s i n A + c o s e c A)^{2} + (c o s A + s e c A)^{2} = 7 + t a n^{2} A + c o t^{2} A

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(s i n A + c o s e c A)^{2} + (c o s A + s e c A)^{2} = 7 + t a n^{2} A + c o t^{2} A .

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\frac{cos A}{1 + sin A} + \frac{1 + sin A}{cos A} = 2 sec A

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Prove the following identity, where the angles involved are acute angles for which the expression is defined. (sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2A

(sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2AL.H.S=(sinA+cosecA)2+(cosA+secA)2=(sin2A+cosec2A+2sinAcosecA)+(cos2A+sec2A+2cosAsecA)=(sin2A+cos2A)+2sinA(1sinA)+2cosA(1cosA)+1+tan2A+1+cot2A=1+2+2+2+tan2A+cot2A=7+tan2A+cot2A=R.H.S

Prove the following identity, where the angles involved are acute angles for which the expression is defined.
$(s i n A + c o s e c A)^{2} + (c o s A + s e c A)^{2} = 7 + t a n^{2} A + c o t^{2} A$