Let BC = s; PC = t Leave tallness of the pinnacle alone AB = h. ∠ABC = θ and ∠APC = 90° – θ (∵ the point of height of the highest point of the pinnacle from two focuses P and B are corresponding) ⇒...
Given: sin θ +2 cos θ = 1 Squaring on the two sides, \[\begin{array}{*{35}{l}} \left( sin\text{ }\theta \text{ }+2\text{ }cos\text{ }\theta \right)2\text{ }=\text{ }1 \\...
Given: \[1+sin2\text{ }\theta \text{ }=\text{ }3\text{ }sin\text{ }\theta \text{ }cos\text{ }\theta \] Isolating L.H.S and R.H.S conditions with sin2 θ, We get, NCERT Exemplar Class 10 Maths Chapter...
Let PR = h meter, be the tallness of the pinnacle. The onlooker is remaining at point Q to such an extent that, the distance between the spectator and pinnacle is QR = (20+x) m, where \[QR\text{...
L.H.S= Since, NCERT Exemplar Class 10 Maths Chapter 8 Ex. 8.4 Question 2 = R.H.S Henceforth, demonstrated.
As per the inquiry, Since, NCERT Exemplar Class 10 Maths Chapter 8 Ex. 8.4 Question 1 Henceforth, demonstrated.
using the formulae: tan (90° – θ) = cot θ tan θ + tan (90° – θ)= tan θ + cot θ we get,
proved
L.H.S: (√3 + 1) (3 – Cot30°) \[=\text{ }\left( \surd 3\text{ }+\text{ }1 \right)\text{ }\left( 3\text{ }\text{ }\surd 3 \right)\text{ }\left[ \because cos\text{ }30{}^\circ =\surd 3 \right]\]...
=> LHS = RHS (PROVED)
Solution: As per the inquiry, tan A = ¾ We know, tan A = opposite/base Along these lines, tan A = 3k/4k Where, Opposite = 3k Base = 4k Utilizing Pythagoras Theorem, (hypotenuse)2 = (perpendicular)2...
= LHS Using the formulae: sec2A – tan2A = 1 sec2A – 1 = tan2A we get, = RHS (PROVED)
=> LHS = RHS (Proved)
False, Defense: L.H.S \[=\text{ }\left( tan\text{ }\theta +2 \right)\text{ }\left( 2\text{ }tan\text{ }\theta +1 \right)\] \[=\text{ }2\text{ }tan2\text{ }\theta \text{ }+\text{ }tan\text{ }\theta...
True: Reason: As indicated by the inquiry, \[cos\text{ }A+cos2\text{ }A\text{ }=\text{ }1\] i.e., \[cos\text{ }A\text{ }=\text{ }1-cos2\text{ }A\] Since, \[sin2\text{ }\theta +cos2\text{ }\theta...
Solution: True, reason:
False, Defense: We realize that, sin θ increments when 0° ≤ θ ≤ 90° cos θ diminishes when 0° ≤ θ ≤ 90° What's more, (sin 80°-cos 80°) = (expanding esteem diminishing worth) = a positive worth....
False, Reason: Since, \[\left( a2-b2 \right)\text{ }=\text{ }\left( a+b \right)\left( a-b \right)\] \[cos2\text{ }23{}^\circ \text{ }\text{ }sin2\text{ }67{}^\circ \text{ }=\left( cos\text{...
1. tan 47o/cot 43 ° = 1 Valid Legitimization: Since, tan (90° - θ) = bed θ
(A) 1/√3 (B) √3 (C) 1 (D) 0 (C) 1 As indicated by the inquiry, \[cos\text{ }9\propto =\text{ }sin\propto and\text{ }9\propto <90{}^\circ \] for example 9α is an intense point We realize that,...
(A) 0 (B) 1 (C) 2 (D) ½ (B) 1 tan 1°. tan 2°.tan 3° … tan 89° =tan1°.tan 2°.tan 3°… tan 43°.tan 44°.tan 45°.tan 46°.tan 47°… tan 87°.tan 88°.tan 89° Since, tan 45° = 1, = tan1°.tan 2°.tan 3°… tan...
(A) cos β (B) cos 2β (C) sin α (D) sin 2α (B) cos 2β As indicated by the inquiry, \[cos\left( \alpha +\beta \right)\text{ }=\text{ }0\] Since, cos 90° = 0 We can compose, \[cos\left( \alpha...
(A) b/√(b2– a2) (B) b/a (C) √(b2-a2)/b (D) a/√(b2-a2) (C) √(b2 – a2)/b As indicated by the inquiry, \[sin\text{ }\theta \text{ }=a/b\] We know, \[sin2\text{ }\theta \text{ }+cos2\text{ }\theta...
(A) – 1 (B) 0 (C) 1 (D) 3 2 (B) 0 As indicated by the inquiry, We need to discover the worth of the situation, \[cosec\left( 75{}^\circ +\theta \right)\text{ }\text{ }sec\left( 15{}^\circ...
(A) √3 (B) 1/√3 (C) √3/2 (D) 1 (A) √3 As indicated by the inquiry, \[Sin\text{ }A\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\ldots \text{ }\left( 1 \right)\] We realize that, NCERT...
(A) 3/5 (B) ¾ (C) 4/3 (D) 5/3 (B) 3/4 As indicated by the inquiry, \[cos\text{ }A\text{ }=\text{ }4/5\text{ }\ldots \text{ }\left( 1 \right)\] We know, \[tan\text{ }A\text{ }=\text{ }sinA/cosA\] To...