Question

Guyzz.... pls solve the 30th question. The best ans will be marked as brainliest

Answer 1

secA =p- tanA

squaring both sides: we get

sec2A=(p−tanA)2

sec2A=p2+tan2A−2ptanA

sec2A−tan2A=p2−2ptanA,since we know the identity(1+tan2A=secA)

hence, 1= p2−2ptanA

tanA= (p2−1)/2p

now,according to right angled triangle, tanA= perpendicular(P)/base(b)

and sinA=perpendicular(P)/hypotnuese(h)

hence we know. P= p2−1,base=2p

so, H= p2+1(calculated by hypotnuese theorem)

now SinA= P/H= (p2−1)/(p2+1)−−ans

Answer 2

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$\huge\mathfrak\blue{hello\:frd}$

$\bf{here..is..ur.. answer}$

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$<b>secA =p- tanA$

$<b>squaring both sides: we get$

$<b>sec2A=(p−tanA)2$

$<b>sec2A=p2+tan2A−2ptanA$

$<b>sec2A−tan2A=p2−2ptanA,since we know the identity(1+tan2A=secA)$

$<b>hence, 1= p2−2ptanA$

$<b>tanA= (p2−1)/2p$

$<b>now,according to right angled triangle, tanA= perpendicular(P)/base(b)$

$<b>and sinA=perpendicular(P)/hypotnuese(h)$

$<b>hence we know. P= p2−1,base=2p$

$<b>so, H= p2+1(calculated by hypotnuese theorem)$

$<b> now SinA= P/H= (p2−1)/(p2+1)−−ans$

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$\bf{By-RakeshPatel}$
$\huge\mathfrak\red{thank\:u}$
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