Advertisements
Advertisements
Question
If tan A =` 5/12` , find the value of (sin A+ cos A) sec A.
Solution
(sin A + cos A ) sec A
= `( sinA + cos A ) 1/ cos A`
=`(sinA )/( cos A) + ( cos A)/( cos A)`
= tan A + 1
= `5/12 +1/1`
=` (5+12)/12`
=`17/12`
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities
`cos theta/(1 - sin theta) = (1 + sin theta)/cos theta`
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
Prove the following trigonometric identities.
`(cosec A)/(cosec A - 1) + (cosec A)/(cosec A = 1) = 2 sec^2 A`
Prove the following identities:
`cosA/(1 - sinA) = sec A + tan A`
Write the value of sin A cos (90° − A) + cos A sin (90° − A).
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
If sec θ + tan θ = x, then sec θ =
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that `x^2 + y^2 + z^2 = r^2`
If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)
There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole.
If x = r sin θ cos Φ, y = r sin θ sin Φ and z = r cos θ, prove that x2 + y2 + z2 = r2.
If tan A + sin A = m and tan A - sin A = n, then show that m2 - n2 = 4 `sqrt(mn)`.
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt("a"^2 + "b"^2 -"c"^2)`
Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
