Advertisements
Advertisements
Question
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
Options
tan2 θ
sec2 θ
1
–1
Solution
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is –1.
Explanation:
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`
= `(cos^2 θ - 1)/(sin^2 θ)`
= `(-sin^2 θ)/(sin^2 θ)` ...(∵ sin2θ = 1 – cos2θ)
= –1
RELATED QUESTIONS
Evaluate without using trigonometric tables:
`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
Prove the following identities:
`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
`(sec^2 theta-1) cot ^2 theta=1`
If x = a sin θ and y = bcos θ , write the value of`(b^2 x^2 + a^2 y^2)`
If `sec theta + tan theta = x," find the value of " sec theta`
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
sec4 A − sec2 A is equal to
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a2 – 1) is equal to
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
If sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is ______.
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
