Advertisements
Advertisements
Question
If \[\cos A = \frac{7}{25}\] find the value of tan A + cot A.
Advertisements
Solution
Given: `cos A=7/25`
We know that,
`sin^2 A+cos^2 A=1`
⇒` sin^2 A+(7/25)^2=1`
⇒` sin^2 A+49/625=1`
⇒` sin^2 A1-49/625`
⇒ `sin^2A=576/625`
⇒ `sin A=24/25`
Therefore,
`tan A+cot A= sin A/cos A+cos A/sin A`
=` (24/25)/(7/25)+1=(7/25)/(24/25)`
= `24/7+7/24`
=`((24)^2+(7)^2)/168`
=`(576+49)/168`
=`625/168`
APPEARS IN
RELATED QUESTIONS
Evaluate without using trigonometric tables:
`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`
Prove the following trigonometric identities.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
If `sec theta = x ,"write the value of tan" theta`.
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2 − q2
If a cos θ − b sin θ = c, then a sin θ + b cos θ =
Prove the following identity :
`cosecA + cotA = 1/(cosecA - cotA)`
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Without using trigonometric table , evaluate :
`(sin49^circ/sin41^circ)^2 + (cos41^circ/sin49^circ)^2`
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a2 – 1) is equal to
If `tan θ = 13/12`, then cot θ = ?
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ± 3.
If cosec A – sin A = p and sec A – cos A = q, then prove that `(p^2q)^(2/3) + (pq^2)^(2/3) = 1`.
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`
