Question

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Answer 1

Answer:

q(8) 16 cm2

q(9) cosA

Step-by-step explanation:

Answer 2

Answer:

ans 1) As

$ab || cd$

triangle AOB similar triangle COD

${(\frac{area \: of \: triangle \: aob}{area \: of \: triangle \: cod} )} = ({ \frac{ab}{cd}) }^{2}$

$\frac{64}{area \: of \: triangle \: cod} = ( { \frac{2cd}{cd} )}^{2} \\ = area \: of \: triangle \: cod \: = \frac{64}{4 } \\ = 16$

ans 2)

$(1 - \sin(a)) ( \tan (a)+ \sec(a) ) \\ = \tan(a) + \sec(a) - \sin(a) \tan(a) \\ - \sin(a) \sec(a) \\ = \tan(a) + \frac{1}{ \cos(a) } - \frac{ { \sin(a) }^{2} }{ \cos(a) } \\ - \tan(a) \\ = \frac{1 - { \sin}^{2} (a)}{ \cos(a) } = \frac{ { \cos}^{2} (a)}{ \cos(a) } \\ = \cos(a)$