Question
Proved that $cosec^2(90^\circ - \theta ) - \tan^2 \theta = \cos^2(90^\circ - \theta ) + \cos^2 \theta .$
✓
Answer
$LHS = cosec^2(90^\circ - \theta ) - \tan^2 \theta$
$LHS = sec^2\theta - \tan^2 \theta = 1$
$RHS = \cos^2(90^\circ - \theta ) + \cos^2 \theta$
$RHS = \sin^2\theta + \cos^2 \theta = 1$
Hence, $LHS = RHS$
Hence proved.
$LHS = sec^2\theta - \tan^2 \theta = 1$
$RHS = \cos^2(90^\circ - \theta ) + \cos^2 \theta$
$RHS = \sin^2\theta + \cos^2 \theta = 1$
Hence, $LHS = RHS$
Hence proved.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In the following figures, O is the centre of the circle. Find the values of a, b, c and d.
$A (5, x), B (-4, 3)$ and $C\ (y, -2)$ are the vertices of the $\triangle ABC$ whose centroid is the origin. Calculate the values of $x$ and $y.$
→Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius $r \ cm.$
→A die is thrown once. Find the probability of getting a number : less than 8
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DBA
Also, show that the Δ AOD is an equilateral triangle.
Calculate : ∠DBA
Also, show that the Δ AOD is an equilateral triangle.
Find the distance between the following pairs of point in the coordinate plane :
$(13 , 7)$ and $(4 , -5)$
→$(13 , 7)$ and $(4 , -5)$
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will be a face card of red colour.
Two circles are drawn with sides AB, AC of a triangle ABC as diameters. They intersect at a point D. Prove that D lies on BC.
Find $x$ and $y$ if $\left[\begin{array}{cc}x+y & y \\ 2 x & x-y\end{array}\right]\left[\begin{array}{c}2 \\ -1\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
→Find the sum of first $22$ terms, of an A.P. in which $d = 7$ and $a_{22}$ is $149$.
→