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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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Verified answer
Question:
If [tex] \sf { \sin \theta = \dfrac{3}{4}} [/tex], find [tex] \sf { \cos \theta} [/tex]
Solution:
We are given that,
[tex] \sf { \sin \theta = \dfrac{Perpendicular }{Hypotenuse }} [/tex]
So,
Now, by pythagoras property -
[tex] \sf\red { \qquad {H} ^{2}= {B} ^{2} + {P} ^{2}} [/tex]
[tex] \sf\blue { \qquad \leadsto {B} ^{2}= {H} ^{2} - {P} ^{2}} [/tex]
[tex] \sf { \qquad \leadsto {B} ^{2}= {4} ^{2} - {3} ^{2}} [/tex]
[tex] \sf { \qquad \leadsto {B} ^{2}= 16 - 9} [/tex]
[tex] \sf { \qquad \leadsto {B} ^{2}= 7} [/tex]
[tex] \sf { \qquad \leadsto B= \sqrt{7} } [/tex]
We know that-
Substitute the values -
[tex] \sf\red { \qquad \cos \theta = \dfrac{\sqrt{7}}{4 } ( Ans) } [/tex]
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Question:
Evaluate [tex] \sf { \dfrac{tan65°}{cot25°}} [/tex]
Solution:
[tex] \sf\red { \qquad \dfrac{tan65°}{cot25°}} [/tex]
[tex] \sf { \qquad \leadsto = \dfrac{tan(95°-25°)}{cot25° } } [/tex]
[ since, 25 + 65 = 90 so, 90 - 25 = 65]
[tex] \sf { \qquad \leadsto = \dfrac{cot25°}{cot25° } } [/tex]
[ since, tan( 90- θ) =cot θ, value of θ is here 25°]
[tex] \sf { \qquad \leadsto = \cancel { \dfrac{cot25°}{cot25°} } = 1 ( Ans) } [/tex]
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Question :
What is the value of [tex] \sf {{cos} ^{2}67°- {sin} ^{2}23°} [/tex]?
Solution:
[tex] \sf\red { \qquad {cos} ^{2}67°- {sin} ^{2}23°} [/tex]
[tex] \sf { \qquad \leadsto {cos} ^{2}67°- sin(90°-67°)} [/tex]
[ since, 23 + 67 = 90 so, 90 - 67 = 23]
[tex] \sf { \qquad \leadsto {cos} ^{2}67°-{cos} ^{2}67° } [/tex]
[ since, sin( 90- θ) =cos θ, value of θ is here 67°]
[tex] \sf { \qquad \leadsto 0( Ans) } [/tex]
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Additional Information :
Trigonometric ratios on an angle of ( 90- θ )
Trigonometric ratios on an angle of ( 90 + θ ) -
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