Question

if tanA = 40/9 , then find secA and sinA by using identity


plz solve it​

Answer 1

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Q) If tanA = 40/9 then find the value of secA and sinA using identities .

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★ Given :–

• tanA = $\dfrac{40}{9}$

★ To Find :–

• sec A = ?
• sin A = ?

★ Solution :–

Using the Formula -

$\boxed{ \sf{ {tan}^{2} \: \theta + 1 = {sec}^{2} \: \theta }}$

Put the value of tan a .

$\mapsto \rm \: { \{ \frac{40}{9 } } \}^{2} + 1 = {sec}^{2} \: a$

$\mapsto \rm \: \frac{4 0 \times 40}{9 \times 9} + 1 = {sec}^{2} \: a$

$\mapsto \rm \: \frac{1600}{81} + 1 = {sec}^{2} \: a$

$\mapsto \rm \: \frac{1600 + 81}{81} = {sec}^{2} \: a$

$\mapsto \rm \: \frac{1681}{81} = {sec}^{2} \: a$

$\mapsto \rm \: sec \: a = \sqrt{ \frac{1681}{81} }$

$\rightarrow \underline{ \boxed{ \frak{ \pink{sec \: a }}= \frac{41}{9} }}$

We know ,

$\mapsto \rm \: cos \: a = \frac{1}{sec \: a}$

$\mapsto \rm \: cos \: a = \frac{1}{ \frac{41}{9} }$

$\mapsto \boxed{ \sf{cos \: a = \frac{9}{41} } }$

Now ,

Using the Formula

$\boxed{ \sf{ {sin}^{2} \: \theta + {cos}^{2} \: \theta = 1}}$

put the value of cos

$\mapsto \rm \: {sin}^{2} \: a + {( \frac{9}{41}) }^{2} = 1$

$\mapsto \rm \: {sin}^{2} \: a = 1 - {( \frac{9}{41} )}^{2}$

$\mapsto \rm \: {sin}^{2} \: a = 1 - \frac{81}{1681}$

$\mapsto \rm \: {sin}^{2} \: a = \frac{1681 - 81}{1681}$

$\mapsto \rm \: sin \: a = \sqrt{ \frac{1600}{1681} }$

$\mapsto \underline{ \boxed{ \frak{ \pink{sin \: a }= \frac{40}{41} }}}$

So ,

• sec A = $\dfrac{41}{9}$
• sin A = $\dfrac{40}{41}$

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★ Know More :–

• sin² x + cos² x = 1

• tan² x + 1 = sec² x

• 1 + cot² x = cosec² x

• sin x = $\dfrac{1}{cosec\;x}$

• cos x = $\dfrac{1}{sec\;x}$

• tan x = $\dfrac{1}{cot\;x}$

Answer 2

Answer:

i hope this helps you in solving