We have,cot A = 125Now, tan A = 1cot A = 512Now, sec A = 1 + tan2A = 1 + 25144 = 169144 = 1312Now, cos A = 1sec A = 1213Now, sin A = 1 - cos2A = 1 - 144169 = 513Now, LHS = tan2A - sin2A=5122 - 5132=25144 - 25169=25169 - 144144 × 169=62524336RHS = sin4A sec2A=5134 × 13122=62524336So, LHS = RHS

Pls solve Q130

Pls solve Q130 tan A I + cot B 12 180. If cot A = vcnly tan: A • Sin: A z sin' A • A 131. In a nght tnangie ARC. nght angled at the rat"" of A B to AC I sii , find tl , find the value of

We have, \cot A = \frac{12}{5} Now, \tan A = \frac{1}{\cot A} = \frac{5}{12} Now, \sec A = \sqrt{1 + \tan^{2} A} = \sqrt{1 + \frac{25}{144}} = \sqrt{\frac{169}{144}} = \frac{13}{12} Now, \cos A = \frac{1}{\sec A} = \frac{12}{13} Now, \sin A = \sqrt{1 - \cos^{2} A} = \sqrt{1 - \frac{144}{169}} = \frac{5}{13} Now, LHS = \tan^{2} A - \sin^{2} A = {(\frac{5}{12})}^{2} - {(\frac{5}{13})}^{2} = \frac{25}{144} - \frac{25}{169} = 25 [\frac{169 - 144}{144 \times 169}] = \frac{625}{24336} RHS = \sin^{4} A . \sec^{2} A = {(\frac{5}{13})}^{4} \times {(\frac{13}{12})}^{2} = \frac{625}{24336} So, LHS = RHS

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Pls solve Q130 Pls solve Q130 tan A I + cot B 12 180. If cot A = vcnly tan: A • Sin: A z sin' A • A 131. In a nght tnangie ARC. nght angled at the rat"" of A B to AC I sii , find tl , find the value of

Pls solve Q130

Pls solve Q130 tan A I + cot B 12 180. If cot A = vcnly tan: A • Sin: A z sin' A • A 131. In a nght tnangie ARC. nght angled at the rat"" of A B to AC I sii , find tl , find the value of