If cos(θ+ϕ)=mcos(θ−ϕ) ,then tanθ is equal to a.(1+m1−m)tan φ b. (1−m1+m)tan φ c.(1−m1+m)cot φ d. (1+m1−m)cot φ Video Solution | ShareSave Answer Step by step video & image solution for If cos(theta+phi)=mcos(theta-phi) ,then tantheta is...
Iftanθ=t,thentan2θ+sec2θis equal to (a)1+t1−t(b)1−t1+t(c)2t1−t(d)2t1+t View Solution Solve:cot2θ=tanθ View Solution Ifθand2θ−45∘are acute angles such thatsinθ=cos(2θ−45∘), thentanθis equal to (a) 1 (b)−1(c)√3(d)1√3 ...
The value of tan 180 degrees is equal to 0. Learn how to find the value of tan 180 degrees and how to derive the formula for tan 180 minus theta, here at BYJU’S today!
Whereas \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \] which can be written as \[\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }\] and similarly \[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }\] then we can define tan 90 degree by $\tan 90^\circ = \dfrac{{...
If any individualfactoron the left side of theequationis equal to00, the entireexpressionwill be equal to00. 2tan(θ)−1=02tan(θ)-1=0 tan(θ)+2=0tan(θ)+2=0 Set2tan(θ)−1-equal to00and solve forθθ. Tap for more steps... ...
Assume a right-angled triangle {eq}XYZ {/eq} with {eq}\angle XYZ=90{}^\circ {/eq}. Here, the side {eq}XZ {/eq} is the hypotenuse. Let the angle {eq}\theta {/eq} be equal to {eq}\angle XZY {/eq}. The opposite side is the side opposite to the angle {eq...
To convert an inverse tangent (tan-1) to an inverse sine (sin-1), use the identity tan-1(x) = sin-1(x/√(1+x2)). We can understand this formula by looking at a righttrianglewith an angle theta and the opposite sidexand adjacent side 1. ...
Find cos 2theta if sin theta = 1/5. Assume that theta is greater than or equal to 0 and less than or equal to 2 . If sec (theta) = {26} / {10}, 0 less than or equal to theta less than or equal to pi / 2 then find: a. sin (theta). b. cos (theta). c....
np.tan(np.radians(0.5* np.abs(degto180(qdr%360.- next_qdr%360.)))# Check whether shift based dist is required, set closer than WP turn distanceswreached = np.where(bs.traf.swlnav * ((dist < self.turndist)+circling))[0]# Return True/1.0 for a/c where we have reached waypoint...
np.tan(np.radians(0.5* np.abs(degto180(qdr%360.- next_qdr%360.)))# Check whether shift based dist is required, set closer than WP turn distanceswreached = np.where(bs.traf.swlnav * ((dist < self.turndist)+circling))[0]# Return True/1.0 for a/c where we have reached waypoint...