cosA=2/5 then find value of 4+4tan²A
cos A = 2/5 = 2x/5x
Hypotenuse = 5x
Height = 2x
Now by Pythagoras Theorem
Hypotenuse² = Height² + Base²
5x² = 2x² + Base²
Base² = 5x² - 2x²
Base² = 25x² - 4x²
Base² = 21x²
Base = √21x²
Base = √21x
Now,
tan A = Base/Height = √22x/2x = √21/2
So,
4 + 4 tan²A :
= 4 + 4 × (√21/2)²
= 4 + 4 × 21/4
= 4 + 21
= 25 ans
Answer:
cos A = 2/5
4+4tan *tanA
4+4tan²A
1/sec A = 2/5
sec A = 5/2
4 + 4 tan²A
4( 1 + tan²A)
4 ( sec²A)
4(5/2)²
4 * 25 /4
= 25
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Answers & Comments
cosA=2/5 then find value of 4+4tan²A
cos A = 2/5 = 2x/5x
Hypotenuse = 5x
Height = 2x
Now by Pythagoras Theorem
Hypotenuse² = Height² + Base²
5x² = 2x² + Base²
Base² = 5x² - 2x²
Base² = 25x² - 4x²
Base² = 21x²
Base = √21x²
Base = √21x
Now,
tan A = Base/Height = √22x/2x = √21/2
So,
4 + 4 tan²A :
= 4 + 4 × (√21/2)²
= 4 + 4 × 21/4
= 4 + 21
= 25 ans
Verified answer
Answer:
cos A = 2/5
4+4tan *tanA
4+4tan²A
cos A = 2/5
1/sec A = 2/5
sec A = 5/2
4 + 4 tan²A
4( 1 + tan²A)
4 ( sec²A)
4(5/2)²
4 * 25 /4
= 25
Plz Mark Me Brainlist If It Helpled You