(English Medium) ICSE Class 10 - CISCE Question Bank Solutions

Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.

Prove that sec² (90° - θ) + tan² (90° - θ) = 1 + 2 cot² θ.

Prove that cosec² (90° - θ) + cot² (90° - θ) = 1 + 2 tan² θ.

Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.

Prove that sin( 90° - θ ) sin θ cot θ = cos²θ.

Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0

Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos²θ.

Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.

If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.

Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.

Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.

If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.

If tan α = n tan β, sin α = m sin β, prove that cos² α  = `(m^2 - 1)/(n^2 - 1)`.

If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.

Without using set squares or protractor construct a triangle ABC in which AB = 4 cm, BC = 5 cm and ∠ABC = 120°.
(i) Locate the point P such that ∠BAp = 90° and BP = CP.
(ii) Measure the length of BP.

Prove that (cosec A - sin A)( sec A - cos A) sec² A = tan A.

State and draw the locus of a swimmer maintaining the same distance from a lighthouse.

State and draw the locus of a point equidistant from two given parallel lines.

Construct a Δ ABC, with AB = 6 cm, AC = BC = 9 cm; find a point 4 cm from A and equidistant from B and C.

Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`