Question

yrr jaldi se answer dedo iska please jaldi se pl....​

Answer 1

Step-by-step explanation:

2nd answer:

to find: tan 65° /cot25°

=tan (90°-25°)/cot25

=cot25°/ cot25°

=1

3rd answer:

to find cos^267-sin^223

=cos^267- cos(90-23)

=0

1st answer:

given sin theta=3/4=p/h

we know cos theta= b/h

by phythagores

ab^2+bc^2=ac^2

4^2=ac^2-3^3

16=ac^2-9

7=ac^2

√7=ac

so cos theta=√7/4

Hope it will help u...

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Answer 2

Question:

If $\sf { \sin \theta = \dfrac{3}{4}}$, find $\sf { \cos \theta}$

Solution:

We are given that,

• $\sf { \sin \theta = \dfrac{3}{4}}$

$\sf { \sin \theta = \dfrac{Perpendicular }{Hypotenuse }}$

So,

• $\sf { Perpendicular =3}$
• $\sf { Hypotenuse = 4}$

Now, by pythagoras property -

$\sf\red { \qquad {H} ^{2}= {B} ^{2} + {P} ^{2}}$

$\sf\blue { \qquad \leadsto {B} ^{2}= {H} ^{2} - {P} ^{2}}$

$\sf { \qquad \leadsto {B} ^{2}= {4} ^{2} - {3} ^{2}}$

$\sf { \qquad \leadsto {B} ^{2}= 16 - 9}$

$\sf { \qquad \leadsto {B} ^{2}= 7}$

$\sf { \qquad \leadsto B= \sqrt{7} }$

We know that-

• $\sf { \cos \theta = \dfrac{Base}{Hypotenuse }}$

Substitute the values -

$\sf\red { \qquad \cos \theta = \dfrac{\sqrt{7}}{4 } ( Ans) }$

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Question:

Evaluate $\sf { \dfrac{tan65°}{cot25°}}$

Solution:

$\sf\red { \qquad \dfrac{tan65°}{cot25°}}$

$\sf { \qquad \leadsto = \dfrac{tan(95°-25°)}{cot25° } }$

[ since, 25 + 65 = 90 so, 90 - 25 = 65]

$\sf { \qquad \leadsto = \dfrac{cot25°}{cot25° } }$

[ since, tan( 90- θ) =cot θ, value of θ is here 25°]

$\sf { \qquad \leadsto = \cancel { \dfrac{cot25°}{cot25°} } = 1 ( Ans) }$

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Question :

What is the value of $\sf {{cos} ^{2}67°- {sin} ^{2}23°}$?

Solution:

$\sf\red { \qquad {cos} ^{2}67°- {sin} ^{2}23°}$

$\sf { \qquad \leadsto {cos} ^{2}67°- sin(90°-67°)}$

[ since, 23 + 67 = 90 so, 90 - 67 = 23]

$\sf { \qquad \leadsto {cos} ^{2}67°-{cos} ^{2}67° }$

[ since, sin( 90- θ) =cos θ, value of θ is here 67°]

$\sf { \qquad \leadsto 0( Ans) }$

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Additional Information :

Trigonometric ratios on an angle of ( 90- θ )

• $\sf {\sin(90- \theta) = \cos \theta}$
• $\sf {\cos(90- \theta) = \sin \theta}$
• $\sf {\tan(90- \theta) = \cot \theta}$
• $\sf {\cot(90- \theta) = \tan \theta}$
• $\sf {\sec(90- \theta) = \cosec \theta}$
• $\sf {\cosec(90- \theta) = \sec \theta}$

Trigonometric ratios on an angle of ( 90 + θ ) -

• $\sf {\sin(90 + \theta) = \cos \theta}$
• $\sf {\cos(90 + \theta) = - \sin \theta}$
• $\sf {\tan(90 +\theta) = - \cot \theta}$
• $\sf {\cot(90 +\theta) = - \tan \theta}$
• $\sf {\sec(90 + \theta) = - \cosec \theta}$
• $\sf {\cosec(90+ \theta) = \sec \theta}$

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