Advertisements
Advertisements
Question
If sin θ + cos θ = `sqrt(3)`, then show that tan θ + cot θ = 1
Advertisements
Solution
sin θ + cos θ = `sqrt(3)` ......[Given]
∴ (sin θ + cos θ)2 = 3 ......[Squaring on both sides]
∴ sin2θ + 2sinθ cosθ + cos2θ = 3 ......[∵ (a + b)2 = a2 + 2ab + b2]
∴ (sin2θ + cos2θ) + 2sinθ cosθ = 3
∴ 1 + 2 sin θ cos θ = 3 ......[∵ sin2θ + cos2θ = 1]
∴ 2 sin θ cos θ = 2
∴ sin θ cos θ = 1 ......(i)
tan θ + cot θ = `sintheta/costheta + costheta/sintheta`
= `(sin^2theta + cos^2theta)/(costhetasintheta)`
= `1/(sintheta costheta)` ......[∵ sin2θ + cos2θ = 1]
= `1/1` ......[From (i)]
= 1
APPEARS IN
RELATED QUESTIONS
Prove that `sqrt(sec^2 theta + cosec^2 theta) = tan theta + cot theta`
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Show that : tan 10° tan 15° tan 75° tan 80° = 1
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
cosec4θ − cosec2θ = cot4θ + cot2θ
What is the value of 9cot2 θ − 9cosec2 θ?
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
\[\frac{x^2 - 1}{2x}\] is equal to
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that `x^2 + y^2 + z^2 = r^2`
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Find the value of sin 30° + cos 60°.
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
Prove that: sin4 θ + cos4θ = 1 - 2sin2θ cos2 θ.
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
