{"id":7774,"date":"2022-09-18T05:45:58","date_gmt":"2022-09-18T05:45:58","guid":{"rendered":"https:\/\/mdr.foobrdigital.com\/?p=7774"},"modified":"2022-09-18T05:45:58","modified_gmt":"2022-09-18T05:45:58","slug":"jee-main-maths-trig-previous-year-questions-with-sol","status":"publish","type":"post","link":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/2022\/09\/18\/jee-main-maths-trig-previous-year-questions-with-sol\/","title":{"rendered":"JEE Main Maths Trig Previous Year Questions With Sol."},"content":{"rendered":"\n

Question 1:<\/strong> The general solution of sin x \u2212 3 sin2x + sin3x = cos x \u2212 3 cos2x + cos3x is _________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

sinx \u2212 3 sin2x + sin3x = cosx \u2212 3 cos2x + cos3x<\/p>\n\n\n\n

\u21d2 2 sin2x cosx \u2212 3 sin2x \u2212 2 cos2x cosx + 3 cos2x = 0<\/p>\n\n\n\n

\u21d2 sin2x (2cosx \u2212 3) \u2212 cos2x (2 cosx \u2212 3) = 0<\/p>\n\n\n\n

\u21d2 (sin2x \u2212 cos2x) (2 cosx \u2212 3) = 0<\/p>\n\n\n\n

\u21d2 sin2x = cos2x<\/p>\n\n\n\n

\u21d2 tan 2x = 1<\/p>\n\n\n\n

2x = n\u03c0 + (\u03c0 \/ 4 )<\/p>\n\n\n\n

x = n\u03c0 \/ 2 + \u03c0 \/ 8<\/p>\n\n\n\n

Question 2:<\/strong> If sec 4\u03b8 \u2212 sec 2\u03b8 = 2, then the general value of \u03b8 is __________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

sec 4\u03b8 \u2212 sec 2\u03b8 = 2 \u21d2 cos 2\u03b8 \u2212 cos 4\u03b8 = 2 cos 4\u03b8 cos 2\u03b8<\/p>\n\n\n\n

\u21d2 \u2212cos 4\u03b8 = cos 6\u03b8<\/p>\n\n\n\n

\u21d2 2 cos 5\u03b8 cos \u03b8 = 0<\/p>\n\n\n\n

When cos 5\u03b8 = 0, 5\u03b8 = (2n + 1)\u03c0\/2<\/p>\n\n\n\n

So \u03b8  = n\u03c0\/5 + \u03c0\/10<\/p>\n\n\n\n

= (2n + 1)\u03c0\/10<\/p>\n\n\n\n

When cos \u03b8 = 0, \u03b8 = (2n+1)\u03c0\/2.<\/p>\n\n\n\n

Question 3:<\/strong> If tan (cot x) = cot (tan x), then sin 2x = ___________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

tan (cot x) = cot (tan x) \u21d2 tan (cot x) = tan (\u03c0 \/ 2 \u2212 tan x)<\/p>\n\n\n\n

cot x = n\u03c0 + \u03c0 \/ 2 \u2212 tanx<\/p>\n\n\n\n

\u21d2 cot x + tan x = n\u03c0 + \u03c0 \/ 2<\/p>\n\n\n\n

1\/sin x cos x = n\u03c0 + \u03c0 \/ 2<\/p>\n\n\n\n

1\/sin 2x = n\u03c0\/2  + \u03c0 \/ 4<\/p>\n\n\n\n

\u21d2 sin2x = 2 \/ [n\u03c0 + {\u03c0 \/ 2}]<\/p>\n\n\n\n

= 4 \/ {(2n + 1) \u03c0}<\/p>\n\n\n\n

Question 4:<\/strong> If the solution for \u03b8 of cosp\u03b8 + cosq\u03b8 = 0, p > 0, q > 0 are in A.P., then numerically the smallest common difference of A.P. is ___________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given cosp\u03b8 = \u2212cosq\u03b8 = cos (\u03c0 + q\u03b8)<\/p>\n\n\n\n

p\u03b8 = 2n\u03c0 \u00b1 (\u03c0 + q\u03b8), n \u2208 I<\/p>\n\n\n\n

\u03b8 = [(2n + 1)\u03c0] \/ [p \u2212 q] or [(2n \u2212 1)\u03c0] \/ [p + q], n \u2208 I<\/p>\n\n\n\n

Both the solutions form an A.P. \u03b8 = [(2n + 1)\u03c0] \/ [p \u2212 q] gives us an A.P. with common difference 2\u03c0 \/ [p \u2212 q] and \u03b8 = [(2n \u2212 1)\u03c0] \/ [p + q] gives us an A.P. with common difference = 2\u03c0 \/ [p + q].<\/p>\n\n\n\n

Certainly, {2\u03c0 \/ [p + q]} < {\u22232\u03c0 \/ [p \u2212 q]\u2223}.<\/p>\n\n\n\n

Question 5:<\/strong> If \u03b1 , \u03b2 are different values of x satisfying a cosx + b sinx = c, then tan ([\u03b1 + \u03b2] \/ 2) = ______________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

a cosx + b sinx = c \u21d2 a {[(1 \u2212 tan2 <\/sup>(x \/ 2)] \/ [1 + tan2 <\/sup>(x \/ 2)]} + 2b {[tan (x \/ 2) \/ 1 + tan2 <\/sup>(x \/ 2)} = c<\/p>\n\n\n\n

\u21d2 (a + c) * tan2 <\/sup>[x \/ 2] \u2212 2b tan [x \/ 2] + (c \u2212 a) = 0<\/p>\n\n\n\n

This equation has roots tan [\u03b1 \/ 2] and tan [\u03b2 \/ 2].<\/p>\n\n\n\n

Therefore, tan [\u03b1 \/ 2] + tan [\u03b2 \/ 2] = 2b \/ [a + c] and tan [\u03b1 \/ 2] * tan [\u03b2 \/ 2]<\/p>\n\n\n\n

= [c \u2212 a] \/ [a + c]<\/p>\n\n\n\n

Now<\/p>\n\n\n\n

tan ((\u03b1 + \u03b2)\/2) = {tan [\u03b1 \/ 2] + tan [\u03b2 \/ 2]} \/ {1 \u2212 tan [\u03b1 \/ 2] * tan [\u03b2 \/ 2]}<\/p>\n\n\n\n

= {[2b] \/ [a + c]} \/ {1\u2212 ([c \u2212 a] \/ [a + c])}<\/p>\n\n\n\n

= b\/a<\/p>\n\n\n\n

Question 6: <\/strong>In a triangle, the length of the two larger sides are 10 cm and 9 cm, respectively. If the angles of the triangle are in arithmetic progression, then the length of the third side in cm can be _________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that in a triangle larger the side, larger the angle.<\/p>\n\n\n\n

Since angles \u2220A, \u2220B and \u2220C are in AP.<\/p>\n\n\n\n

Hence, \u2220B = 60o<\/sup> cosB = [a2<\/sup> + c2<\/sup> \u2212b2<\/sup>] \/ [2ac]<\/p>\n\n\n\n

\u21d2 1 \/ 2 = [100 + a2 <\/sup>\u2212 81] \/ [20a]<\/p>\n\n\n\n

\u21d2 a2<\/sup> + 19 = 10a<\/p>\n\n\n\n

\u21d2 a2 <\/sup>\u2212 10a + 19 = 0<\/p>\n\n\n\n

a = 10 \u00b1 (\u221a[100 \u2212 76] \/ [2])<\/p>\n\n\n\n

\u21d2 a = 5 \u00b1 \u221a6<\/p>\n\n\n\n

Question 7: <\/strong>In triangle ABC, if \u2220A = 45\u2218, \u2220B = 75\u2218, then a + c\u221a2 = __________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220C = 180o<\/sup> \u2212 45o<\/sup> \u2212 75o<\/sup> = 60o<\/sup><\/p>\n\n\n\n

a\/sin A = b\/sin b = c\/sin C<\/p>\n\n\n\n

a\/sin 45 = b\/sin 75 = c\/sin 60<\/p>\n\n\n\n

=> \u221a2a = 2\u221a2b\/(\u221a3+1) = 2c\/\u221a3<\/p>\n\n\n\n

=> a = 2b\/(\u221a3+1)<\/p>\n\n\n\n

c = \u221a6b\/(\u221a3+1)<\/p>\n\n\n\n

a+\u221a2c = [2b\/(\u221a3+1)] + [\u221a12b\/(\u221a3+1)]<\/p>\n\n\n\n

Solving, we get<\/p>\n\n\n\n

= 2b<\/p>\n\n\n\n

Question 8:<\/strong> If cos\u22121 <\/sup>p + cos\u22121<\/sup> q + cos\u22121<\/sup> r = \u03c0 then p2<\/sup> + q2<\/sup> + r2<\/sup> + 2pqr = ________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given cos\u22121 <\/sup>p + cos\u22121<\/sup> q + cos\u22121<\/sup> r = \u03c0<\/p>\n\n\n\n

cos\u22121 <\/sup>p + cos\u22121<\/sup> q = \u03c0 \u2013 cos\u22121<\/sup> r<\/p>\n\n\n\n

cos\u22121 <\/sup>(pq \u2013 \u221a(1 \u2013 p2<\/sup>) \u221a(1 \u2013 q2<\/sup>) = cos-1<\/sup> (-r)<\/p>\n\n\n\n

(pq \u2013 \u221a(1 \u2013 p2<\/sup>) \u221a(1 \u2013 q2<\/sup>) = -r<\/p>\n\n\n\n

(pq + r) = \u221a(1 \u2013 p2<\/sup>) \u221a(1 \u2013 q2<\/sup>)<\/p>\n\n\n\n

squaring<\/p>\n\n\n\n

(pq + r)2<\/sup> = (1 \u2013 p2<\/sup>) (1 \u2013 q2<\/sup>)<\/p>\n\n\n\n

p2<\/sup>q2<\/sup> + 2pqr + r2<\/sup> = 1 \u2013 p2<\/sup> \u2013 q2<\/sup> + p2<\/sup>q2<\/sup><\/p>\n\n\n\n

p2<\/sup> + q2<\/sup> + r2<\/sup> + 2pqr = 1<\/p>\n\n\n\n

Question 9:<\/strong> tan [(\u03c0 \/ 4) + (1 \/ 2) * cos\u22121 <\/sup>(a \/ b)] + tan [(\u03c0 \/ 4) \u2212 (1 \/ 2) cos\u22121 .<\/sup>(a \/ b)] = _________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

tan [(\u03c0 \/ 4) + (1 \/ 2) * cos\u22121 <\/sup>(a \/ b)] + tan [(\u03c0 \/ 4) \u2212 (1 \/ 2) cos\u22121 .<\/sup>(a \/ b)]<\/p>\n\n\n\n

Let (1 \/ 2) * cos\u22121 <\/sup>(a \/ b) = \u03b8<\/p>\n\n\n\n

\u21d2 cos 2\u03b8 = a \/ b<\/p>\n\n\n\n

Thus, tan [{\u03c0 \/ 4} + \u03b8] + tan [{\u03c0 \/ 4} \u2212 \u03b8] = [(1 + tan\u03b8) \/ (1 \u2212 tan\u03b8)]+ ([1 \u2212 tan\u03b8] \/ [1 + tan\u03b8])<\/p>\n\n\n\n

= [(1 + tan\u03b8)2 <\/sup>+ (1 \u2212 tan\u03b8)2<\/sup>] \/ [(1 \u2212 tan2<\/sup>\u03b8)]<\/p>\n\n\n\n

= [1 + tan2<\/sup>\u03b8 + 2tan\u03b8 + 1 + tan2<\/sup>\u03b8 \u2212 2tan\u03b8] \/ [(1 \u2013 tan2<\/sup>\u03b8)]<\/p>\n\n\n\n

= 2 (1 + tan2<\/sup>\u03b8) \/ [(1 \u2013 tan2<\/sup>\u03b8)]<\/p>\n\n\n\n

= 2 sec2<\/sup>\u03b8 cos2<\/sup>\u03b8\/(cos2<\/sup>\u03b8 \u2013 sin2<\/sup>\u03b8)<\/p>\n\n\n\n

= 2 \/cos2\u03b8<\/p>\n\n\n\n

= 2 \/ [a \/ b]<\/p>\n\n\n\n

= 2b \/ a<\/p>\n\n\n\n

Question 10:<\/strong> The number of real solutions of tan\u22121 <\/sup>\u221a[x (x + 1)] + sin\u22121 <\/sup>[\u221a(x2<\/sup> + x + 1)] = \u03c0 \/ 2 is ___________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

tan\u22121 <\/sup>\u221a[x (x + 1)] + sin\u22121 <\/sup>[\u221a(x2<\/sup> + x + 1)] = \u03c0 \/ 2<\/p>\n\n\n\n

tan\u22121 <\/sup>\u221a[x (x + 1)] is defined when x (x + 1) \u2265 0 ..(i)<\/p>\n\n\n\n

sin\u22121 <\/sup>[\u221ax2<\/sup> + x + 1] is defined when 0 \u2264 x (x + 1) + 1 \u2264 1 or x2<\/sup> + x + 1  \u2265 1 ..(ii)<\/p>\n\n\n\n

From (i) and (ii), x (x + 1) = 0 or x = 0 and -1.<\/p>\n\n\n\n

Hence, the number of solution is 2.<\/p>\n\n\n\n

Question 11:<\/strong> What is the value of sin (cot\u22121 <\/sup>x)?<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let cot\u22121 <\/sup>x = \u03b8 \u21d2 x = cot\u03b8<\/p>\n\n\n\n

Now cosec\u03b8 = \u221a[1 + cot2<\/sup>\u03b8] = \u221a[1 + x2<\/sup>]<\/p>\n\n\n\n

Therefore, sin\u03b8 = 1 \/ cosec\u03b8<\/p>\n\n\n\n

= 1 \/ \u221a[1 + x2<\/sup>]<\/p>\n\n\n\n

\u21d2\u03b8 = sin\u22121 <\/sup>{1 \/ \u221a[1 + x2<\/sup>]}<\/p>\n\n\n\n

Hence, sin (cot\u22121 <\/sup>x) = sin (sin\u22121 <\/sup>{1 \/ \u221a[1 + x2<\/sup>]})<\/p>\n\n\n\n

= 1 \/ \u221a[1 + x2<\/sup>]<\/p>\n\n\n\n

= (1 + x2<\/sup>)\u22121\/2<\/sup><\/p>\n\n\n\n

Question 12: <\/strong>sec2 <\/sup>(tan\u22121 <\/sup>2) + cosec2 <\/sup>(cot\u22121<\/sup> 3) = _________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let (tan\u22121 <\/sup>2) = \u03b1<\/p>\n\n\n\n

\u21d2 tan \u03b1 = 2 and cot\u22121<\/sup> 3 = \u03b2<\/p>\n\n\n\n

\u21d2 cot \u03b2 = 3<\/p>\n\n\n\n

sec2 <\/sup>(tan\u22121 <\/sup>2) + cosec2 <\/sup>(cot\u22121<\/sup> 3)<\/p>\n\n\n\n

= sec2 <\/sup>\u03b1 + cosec2<\/sup>\u03b2<\/p>\n\n\n\n

= 1 + tan2<\/sup>\u03b1 + 1 + cot2<\/sup>\u03b2<\/p>\n\n\n\n

= 2 + (2)2<\/sup> + (3)2<\/sup><\/p>\n\n\n\n

= 15<\/p>\n\n\n\n

Question 13:<\/strong> A vertical pole consists of two parts, the lower part being one-third of the whole. At a point in the horizontal plane through the base of the pole and distance 20 meters from it, the upper part of the pole subtends an angle whose tangent is 1 \/ 2. The possible heights of the pole are _________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let H be the height of the pole.<\/p>\n\n\n\n

tan \u03b1 = H\/60<\/p>\n\n\n\n

tan \u03b8 = 1\/2<\/p>\n\n\n\n

tan (a+b) = (tan a + tan b)\/(1-tan a tan b)<\/p>\n\n\n\n

tan (\u03b1 + \u03b8) = H\/20<\/p>\n\n\n\n

=> (tan \u03b1 + tan \u03b8)\/(1 \u2013 tan \u03b1 tan \u03b8)<\/p>\n\n\n\n

H\/20 = [(H\/60) + (\u00bd)]\/[(1 \u2013 H\/120)<\/p>\n\n\n\n

=> H2<\/sup> \u2013 80H + 1200 = 0<\/p>\n\n\n\n

=> (H \u2013 60) (H-20) = 0<\/p>\n\n\n\n

=> H = 60 or H = 20<\/p>\n\n\n\n

Height of the pole can be 60 m or 20 m.<\/p>\n\n\n\n

Question 14:<\/strong> A tower of height b subtends an angle at a point O on the level of the foot of the tower and a distance a from the foot of the tower. If a pole mounted on the tower also subtends an equal angle at O, the height of the pole is ___________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let AB be the tower and PB be the pole .<\/p>\n\n\n\n

Let AB = b and PB = p<\/p>\n\n\n\n

tan \u03b1 = b \/ a<\/p>\n\n\n\n

tan 2\u03b1 = (p+b)\/a<\/p>\n\n\n\n

2 tan \u03b1\/(1 \u2013 tan2<\/sup> \u03b1) = (p+b)\/a<\/p>\n\n\n\n

=> (2b\/a)\/(1 \u2013 b2<\/sup>\/a2<\/sup>) = (p+b)\/a<\/p>\n\n\n\n

=> 2ab\/(a2<\/sup> \u2013 b2<\/sup>) = (p+b)\/a<\/p>\n\n\n\n

\u21d2 [2ba2<\/sup> \u2212 a2<\/sup>b + b3<\/sup>] \/ [a2<\/sup> \u2212 b2<\/sup>] = p<\/p>\n\n\n\n

\u21d2 p = [b * (a2<\/sup> + b2<\/sup>)] \/ [a2<\/sup> \u2212 b2<\/sup>]<\/p>\n\n\n\n

Question 15: <\/strong>A balloon is observed simultaneously from three points A, B and C on a straight road directly under it. The angular elevation at B is twice and at C is thrice that of A. If the distance between A and B is 200 metres and the distance between B and C is 100 metres, then the height of balloon is given by _________.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x = h cot3\u03b1 \u2026..(i)<\/p>\n\n\n\n

(x + 100) = h cot2\u03b1 \u2026\u2026(ii)<\/p>\n\n\n\n

(x + 300) = h cot\u03b1 \u2026\u2026(iii)<\/p>\n\n\n\n

From (i) and (ii), \u2212100 = h (cot3\u03b1 \u2212 cot2\u03b1),<\/p>\n\n\n\n

From (ii) and (iii), \u2212200 = h (cot2\u03b1 \u2212 cot\u03b1),<\/p>\n\n\n\n

100 = h ([sin\u03b1] \/ [sin3\u03b1 * sin2\u03b1]) and 200 = h ([sin\u03b1] \/ [sin2\u03b1 * sin\u03b1]) or<\/p>\n\n\n\n

sin3\u03b1 \/ sin\u03b1 = 200 \/ 100<\/p>\n\n\n\n

\u21d2 sin3\u03b1 \/ sin\u03b1 = 2<\/p>\n\n\n\n

\u21d2 3 sin\u03b1 \u2212 4 sin3<\/sup>\u03b1 \u2212 2 sin\u03b1 = 0<\/p>\n\n\n\n

\u21d2 4 sin3<\/sup>\u03b1 \u2212 sin\u03b1 = 0<\/p>\n\n\n\n

\u21d2 sin\u03b1 = 0 or<\/p>\n\n\n\n

sin2\u03b1 = 1 \/ 4 = sin2 <\/sup>(\u03c0 \/ 6)<\/p>\n\n\n\n

\u21d2 \u03b1 = \u03c0 \/ 6<\/p>\n\n\n\n

Hence, h = 200 * sin [\u03c0 \/ 3]<\/p>\n\n\n\n

= 200* [\u221a3 \/ 2]<\/p>\n\n\n\n

= 100 \u221a3<\/p>\n","protected":false},"excerpt":{"rendered":"

Question 1: The general solution of sin x \u2212 3 sin2x + sin3x = cos x \u2212 3 cos2x + cos3x is _________. Solution: sinx \u2212 3 sin2x + sin3x = cosx \u2212 3 cos2x + cos3x \u21d2 2 sin2x cosx \u2212 3 sin2x \u2212 2 cos2x cosx + 3 cos2x = 0 \u21d2 sin2x (2cosx […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[475],"tags":[],"_links":{"self":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/posts\/7774"}],"collection":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/comments?post=7774"}],"version-history":[{"count":0,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/posts\/7774\/revisions"}],"wp:attachment":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/media?parent=7774"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/categories?post=7774"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/tags?post=7774"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}