Join 60,000+ competitive exam aspirants
or with email
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Join 60,000+ competitive exam aspirants
If cosec2θ=sec(θ−30)°, then find the measure of θ (in degrees).
40
25
45
35
40
Use the property that csc(2θ)=sec(90°−2θ). Equate 90°−2θ=θ−30° to find 3θ=120°, so θ=40°.
cosec 2θ = sec(θ - 30)°
csc(A)=sec(90°−A)
Use the property that csc(2θ)=sec(90°−2θ). Equate 90°−2θ=θ−30° to find 3θ=120°, so θ=40°.
Confusing the identity for complementary angles; remember that csc of an angle equals sec of the complement (90−θ).
Recall Trigonometric Identity
We know that csc(α)=sec(90°−α) for any angle α.
csc(2θ)=sec(90°−2θ)
Substitute into Equation
Replace csc(2θ) in the given equation csc(2θ)=sec(θ−30°) with the identity.
sec(90°−2θ)=sec(θ−30°)
Solve for θ
Equate the angles since the functions are the same: 90°−2θ=θ−30°. Rearranging terms gives 3θ=120°.
θ=3120°=40°
A is correct because the identity csc(2θ)=sec(90°−2θ) leads to 90°−2θ=θ−30°, resulting in θ=40°.
This identity is part of the standard set sin(90−x)=cos(x) and tan(90−x)=cot(x); mastery of these is essential for fast simplification in RRB exams.
Get instant step-by-step AI explanations, full shift Previous Year Papers (SSC GD, RRB, SSC JE), and regional rank analysis on the official Examoogle App.
Free Download • 100K+ Aspirants