Question

Pls answer the 25 th question

Answer 1

solution

given

A+ B = 90°

then , B = 90° - A

forumla used ;

tan A × cot A = 1

tan (90-A) = cot A

solution

refer to attachment

answer

tan ² A

= tan ² (90-B)= cot ² B

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Answer 2

$<font color ="red">$ANSWER:-

$= \frac{ \tan(a) . \tan(90 - a) + tan(a).cot(90 - a) }{ \sin(a) . \: sec(90 - a)} - \frac{ sin(90 - a)}{cos {}^{2} a} \\ = \frac{ \tan(a) \: . \cot(a) + tan(a).tan(a)}{sin(a). \cosec(a) } = \frac{cos {}^{2}(a )}{cos {}^{2}(a) } \\ </p><p>.°.1+tan {}^{2} (a) - 1 = tan {}^{2} (a) \\ </p><p>.°.tan {}^{2} (90 - b) = cot {}^{2} (b)$

$<font color ="green">$HENCE,

SO, FINAL ANSWER IS cot²b.