JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
\(2 \pi-\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65}\right)\) is equal to
- A \(\frac{7 \pi}{4}\)
- B \(\frac{5 \pi}{4}\)
- C \(\frac{3 \pi}{2}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 \pi}{2}\)
Step-by-step Solution
Detailed explanation
\(2 \pi-\left(\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right)\) \(=2 \pi-\left(\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{16}{63}\right)\right)\)…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- \(\max_{0 \leq x \leq \pi}\left(16\sin\left(\dfrac{x}{2}\right)\cos^3\left(\dfrac{x}{2}\right)\right)\) is equal to:JEE Mains 2026 Medium
- Let \(f: R \rightarrow R\) be a function defined as \(f(x)=\left\{\begin{array}{cl}\frac{\sin (a+1) x+\sin 2 x}{2 x} & , \text { if } x<0 \\ b & , \text { if } x=0 \\ \frac{\sqrt{x+b x^{3}}-\sqrt{x}}{b x^{5 / 2}} & , \text { if } x>0\end{array}\right.\) . If \(f\) is continuous at \(x=0,\) then the value of \(a + b\) is equal to ....... .JEE Mains 2021 Hard
- The set of all values of \(\mathrm{k}\,>\,-1\), for which the equation \(\left(3 x^{2}+4 x+3\right)^{2}-(k+1)\left(3 x^{2}+4 x+3\right)\) \(\left(3 x^{2}+4 x+2\right)+k\left(3 x^{2}+4 x+2\right)^{2}=0\) has real roots is:JEE Mains 2021 Hard
- Let the mean and variance of \(12\) observations be \(\frac{9}{2}\) and \(4\) respectively. Later on, it was observed that two observations were considered as \(9\) and \(10\) instead of \(7\) and \(14\) respectively. If the correct variance is \(\frac{m}{n}\), where \(m\) and \(n\) are co-prime, then \(m + n\) is equal toJEE Mains 2023 Hard
- \(\max _{0 \leq x \leq \pi}\left\{x-2 \sin x \cos x+\frac{1}{3} \sin 3 x\right\}=\)JEE Mains 2023 Hard
- Let \(f\left( x \right) = \int\limits_0^x {g\left( t \right)dt} \), where \(g\) is a non zero even function. If \(f(x+5) = g(x)\) , then \(\int\limits_0^x {f\left( t \right)dt} \) equalsJEE Mains 2019 Hard
More PYQs from JEE Mains
- An urn contains \(5\) red and \(2\) green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, isJEE Mains 2019 Hard
- Let \(A =\left(\begin{array}{cc}2 & -1 \\ 0 & 2\end{array}\right)\). If \(B = I -{ }^{5} C _{1} (\operatorname{adj} A )+{ }^{5} C _{2}\) \((\operatorname{adjA})^{2}-\ldots-{ }^{5} C _{5} (\operatorname{adj} A )^{5}\), then the sum of all elements of the matrix \(B\) isJEE Mains 2022 Medium
- The angle of elevation of the top \(P\) of a tower from the feet of one person standing due South of the tower is \(45^{\circ}\) and from the feet of another person standing due west of the tower is \(30^{\circ}\). If the height of the tower is \(5\) meters, then the distance (in meters) between the two persons is equal to \(..........\).JEE Mains 2023 Medium
- Let the coefficients of \(x ^{-1}\) and \(x ^{-3}\) in the expansion of \(\left(2 x^{\frac{1}{5}}-\frac{1}{x^{\frac{1}{5}}}\right)^{15}, x>0\), be \(m\) and \(n\) respectively. If \(r\) is a positive integer such \(m n^{2}={ }^{15} C _{ r } .2^{ r }\), then the value of \(r\) is equal toJEE Mains 2022 Medium
- If two distinct point \(Q , R\) lie on the line of intersection of the planes \(- x +2 y - z =0\) and \(3 x-5 y+2 z=0\) and \(P Q=P R=\sqrt{18}\) where the point \(P\) is \((1,-2,3)\), then the area of the triangle \(PQR\) is equal toJEE Mains 2022 Hard
- Let \(C_1\) be the circle in the third quadrant of radius 3 , that touches both coordinate axes. Let \(\mathrm{C}_2\) be the circle with centre \((1,3)\) that touches \(\mathrm{C}_1\) externally at the point \((\alpha, \beta)\). If \((\beta-\alpha)^2=\frac{m}{n}, \operatorname{gcd}(m, n)=1\), then \(m+n\) is equal to :JEE Mains 2025 Medium